catatanmatematika.com sarana belajar matematika secara online dan mandiri
Takut Akan Tuhan Adalah Permulaan Pengetahuan
Home / Barisan dan Deret

Kumpulan Soal - Notasi Sigma + Pembahasan

Post a Comment
Hallo...! Pengunjung setia Catatan Matematika, kali ini Bang RP (Reikson Panjaitan, S.Pd) berbagi Kumpulan Soal Notasi Sigma dan Pembahasannya. Ayo... manfaatkan website Catatan Matematika ini untuk belajar matematika secara online.
Tata Cara Belajar:
Cobalah mengerjakan soal-soal yang tersedia secara mandiri. Setelah itu cek jawaban kamu dengan pembahasan yang telah disediakan, dengan cara:
klik "LIHAT/TUTUP:".
Soal Notasi Sigma No. 1
Nilai dari $\sum\limits_{n=3}^{6}{(12-n^2)}$ = ….
A. $-41$
B. $-38$
C. $-35$
D. $-30$
E. $-24$
Penyelesaian: Lihat/Tutup $\sum\limits_{n=3}^{6}{(12-n^2)}$
= $(12-3^2)+(12-4^2)+(12-5^2)+(12-6^2)$
= $3+(-4)+(-13)+(-24)$
= $-38$
Jawaban: B
Soal Notasi Sigma No. 2
Nilai dari $\sum\limits_{p=2}^{5}{\frac{p}{p+1}}$ = ….
A. 3,05
B. 3,10
C. 3,20
D. 3,25
E. 3,50
Penyelesaian: Lihat/Tutup $\sum\limits_{p=2}^{5}{\frac{p}{p+1}}$
= $\frac{2}{2+1}+\frac{3}{3+1}+\frac{4}{4+1}+\frac{5}{5+1}$
= $\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+\frac{5}{6}$
= $\frac{2\times 20}{3\times 20}+\frac{3\times 15}{4\times 15}+\frac{4\times 12}{5\times 12}+\frac{5\times 10}{6\times 10}$
= $\frac{40+45+48+50}{60}$
= 3,05
Jawaban: A
Soal Notasi Sigma No. 3
Notasi sigma yang menyatakan penjumlahan 20 bilangan asli ganjil adalah ….
A. $\sum\limits_{n=1}^{20}{1}$
B. $\sum\limits_{n=1}^{20}{(n-1}$
C. $\sum\limits_{n=1}^{20}{(2n-1)}$
D. $\sum\limits_{n=1}^{10}{(n-1)}$
E. $\sum\limits_{n=1}^{10}{(2n-1)}$
Penyelesaian: Lihat/Tutup $\underbrace{1+3+5+7+...}_{20\,bilangan\,ganjil\,pertama}$
= $(2.1-1)+(2.2-1)+(2.3-1)+...+(2.20-1)$
= $\sum\limits_{n=1}^{20}{(2n-1)}$
Jawaban: C
Soal Notasi Sigma No. 4
Notasi sigma yang menyatakan penjumlahan 25 bilangan kuadrat pertama adalah ….
A. $\sum\limits_{n=1}^{5}{2n}$
B. $\sum\limits_{n=1}^{5}{n^2}$
C. $\sum\limits_{n=1}^{25}{2n}$
D. $\sum\limits_{n=1}^{25}{n^2}$
E. $\sum\limits_{n=1}^{625}{n^2}$
Penyelesaian: Lihat/Tutup $\underbrace{1^2+2^2+3^2+...}_{25\,bilangan\,kuadrat\,pertama}$
= $1^2+2^2+3^2+...+25^2$
= $\sum\limits_{n=1}^{25}{n^2}$
Jawaban: D
Soal Notasi Sigma No. 5
Diketahui $\sum\limits_{i=1}^{10}{a_i}=40$ dan $\sum\limits_{i=1}^{10}{b_i}=50$, maka $\sum\limits_{i=1}^{10}{(4a_i-b_i+2)}$ = ….
A. 100
B. 110
C. 120
D. 130
E. 140
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{i=1}^{10}{(4a_i-b_i+2)} &= 4\sum\limits_{i=1}^{10}{a_i}-\sum\limits_{i=1}^{10}{b_i}+\sum\limits_{i=1}^{10}{2} \\ &= 4.40-50+10.2 \\ &= 160-50+20 \\ &= 130 \end{align}$
Jawaban: D
Soal Notasi Sigma No. 6
Diketahui $\sum\limits_{k=5}^{25}{(2-pk)}=0$, maka nilai $\sum\limits_{k=5}^{25}{pk}$ = ….
A. 20
B. 28
C. 30
D. 42
E. 112
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{k=5}^{25}{(2-pk)} &= 0 \\ \sum\limits_{k=5}^{25}{2}-\sum\limits_{k=5}^{25}{pk} &= 0 \\ \sum\limits_{k=5}^{25}{2} &= \sum\limits_{k=5}^{25}{pk} \\ \sum\limits_{k=5-4}^{25-4}{2} &= \sum\limits_{k=5}^{25}{pk} \\ \sum\limits_{1}^{21}{2} &= \sum\limits_{k=5}^{25}{pk} \\ 21\times 2 &= \sum\limits_{k=5}^{25}{pk} \\ 42 &= \sum\limits_{k=5}^{25}{pk} \end{align}$
Jawaban: D
Soal Notasi Sigma No. 7
Jika bentuk $\sum\limits_{x=4}^{20}{(3-4x)}$ diubah dengan notasi sigma lain yang sesuai dengan batas bawah satu, akan diperoleh bentuk ….
A. $\sum\limits_{x=1}^{20}{(9-4x)}$
B. $\sum\limits_{x=1}^{20}{(3-4x)}$
C. $\sum\limits_{x=1}^{20}{(-9-4x)}$
D. $\sum\limits_{x=1}^{17}{(9-4x)}$
E. $\sum\limits_{x=1}^{17}{(-9-4x)}$
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{x=4}^{20}{(3-4x)} &= \sum\limits_{x=4-3}^{20-3}{(3-4(x+3))} \\ &= \sum\limits_{x=1}^{17}{(3-4x-12)} \\ &= \sum\limits_{x=1}^{17}{(-9-4x)} \end{align}$
Jawaban: E
Soal Notasi Sigma No. 8
Jika batas bawah diubah menjadi tujuh, bentuk notasi sigma $\sum\limits_{n=3}^{7}{(6-3n)}$ akan menjadi ….
A. $\sum\limits_{n=7}^{11}{(6-3n)}$
B. $\sum\limits_{n=7}^{11}{(9-3n)}$
C. $\sum\limits_{n=7}^{11}{(18-3n)}$
D. $\sum\limits_{n=7}^{11}{(24-3n)}$
E. $\sum\limits_{n=7}^{11}{(36-3n)}$
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{n=3}^{7}{(6-3n)} &= \sum\limits_{n=3+4}^{7+4}{(6-3(n-4))} \\ &= \sum\limits_{n=7}^{11}{(6-3n+12)} \\ &= \sum\limits_{n=7}^{11}{(18-3n)} \end{align}$
Jawaban: C
Soal Notasi Sigma No. 9
$\sum\limits_{i=1}^{n}{({{i}^{2}}+5i)}$ = ….
A. $\frac{n(n+1)(n+3)}{6}$
B. $\frac{n(n+1)(n+6)}{6}$
C. $\frac{n(n+1)(2n+3)}{6}$
D. $\frac{n(n+1)(2n+12)}{6}$
E. $\frac{n(n+1)(2n+16)}{6}$
Penyelesaian: Lihat/Tutup $\sum\limits_{i=1}^{n}{({{i}^{2}}+5i)}$
= $\sum\limits_{i=1}^{n}{{{i}^{2}}}+5\sum\limits_{i=1}^{n}{i}$
= $\frac{n(n+1)(2n+1)}{6}+5.\frac{n(n+1)}{2}$
= $\frac{n(n+1)(2n+1)}{6}+\frac{15n(n+1)}{6}$
= $\frac{n(n+1)(2n+1)+15n(n+1)}{6}$
= $\frac{n(n+1)[2n+1+15]}{6}$
= $\frac{n(n+1)(2n+16)}{6}$
Jawaban: E
Soal Notasi Sigma No. 10
Jika $\sum\limits_{i=1}^{n}{(2i+3)}=60$ maka $n$ = ….
A. 4
B. 5
C. 6
D. 7
E. 8
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{i=1}^{n}{(2i+3)} &= 60 \\ 2\sum\limits_{i=1}^{n}{i}+\sum\limits_{i=1}^{n}{3} &= 60 \\ 2.\frac{n(n+1)}{2}+n.3 &= 60 \\ n(n+1)+3n &= 60 \\ n^2+n+3n-60 &= 0 \\ n^2+4n-60 &= 0 \\ (n+10)(n-6) &= 0 \end{align}$
$n+10=0\to n=-10$ (tidak memenuhi)
$n-6=0\to n=6$ (memenuhi)
Jawaban: C
Soal Notasi Sigma No. 11
$\sum\limits_{k=1}^{40}{3k}+\sum\limits_{k=1}^{40}{(k+5)}$ = ….
A. 3460
B. 3465
C. 3470
D. 3475
E. 3480
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{k=1}^{40}{3k}+\sum\limits_{k=1}^{40}{(k+5)} &= \sum\limits_{k=1}^{40}{(3k+(k+5))} \\ &= \sum\limits_{k=1}^{40}{(4k+5)} \\ &= 4\sum\limits_{k=1}^{40}{k}+\sum\limits_{k=1}^{40}{5} \\ &= 4.\frac{40(40+1)}{2}+40.5 \\ &= 3480 \end{align}$
Jawaban: E
Soal Notasi Sigma No. 12
Nilai dari $\sum\limits_{k=1}^{110}{2k}+\sum\limits_{k=1}^{110}{(k+1)}$ adalah ….
A. 37290
B. 36850
C. 18645
D. 18425
E. 18420
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{k=1}^{110}{2k}+\sum\limits_{k=1}^{110}{(k+1)} &= \sum\limits_{k=1}^{110}{(2k+k+1)} \\ &= \sum\limits_{k=1}^{110}{(3k+1)} \\ &= 3\sum\limits_{k=1}^{110}{k}+\sum\limits_{k=1}^{110}{1} \\ &= 3.\frac{110(110+1)}{2}+110 \\ &= 18425 \end{align}$
Jawaban: D
Soal Notasi Sigma No. 13
Jika $p=\sum\limits_{n=1}^{\infty }{\frac{n}{3^n}}$, maka nilai dari $4p$ = ….
A. 1
B. 2
C. 3
D. 4
E. 6
Penyelesaian: Lihat/Tutup $p=\sum\limits_{n=1}^{\infty }{\frac{n}{3^n}}\,....\,(1)$
Dari persamaan (1):
$\begin{align}p &= 3\sum\limits_{n=1}^{\infty }{\frac{n}{3^{n+1}}} \\ \frac{1}{3}p &= \sum\limits_{n=1}^{\infty }{\frac{n}{3^{n+1}}} \\ &= \sum\limits_{n=1+1}^{\infty +1}{\frac{n}{{{3}^{(n-1)+1}}}} \\ \frac{p}{3} &= \sum\limits_{n=2}^{\infty }{\frac{n-1}{3^n}}\,....\,(2) \end{align}$
Persamaan (1) dikurangkan dengan persamaan (2) maka:
$\begin{align}p-\frac{1}{3}p &= \sum\limits_{n=1}^{\infty }{\frac{n}{3^n}}-\sum\limits_{n=2}^{\infty }{\frac{n-1}{3^n}} \\ \frac{2}{3}p &= \frac{1}{3}+\sum\limits_{n=2}^{\infty }{\frac{n}{3^n}}-\sum\limits_{n=2}^{\infty }{\frac{n-1}{3^n}} \\ &= \frac{1}{3}+\sum\limits_{n=2}^{\infty }{\frac{1}{3^n}} \\ &= \frac{1}{3}+\left( \frac{1}{9}+\frac{1}{27}+\frac{1}{81}+... \right) \\ &= \frac{1}{3}+\frac{1/9}{1-\frac{1}{3}} \\ &= \frac{1}{3}+\frac{1}{9}.\frac{3}{2} \\ &= \frac{1}{3}+\frac{1}{6} \\ \frac{2}{3}p &= \frac{1}{2} \\ p &= \frac{1}{2}\times \frac{3}{2} \\ p &= \frac{3}{4} \end{align}$
$4p=4.\frac{3}{4}=3$
Jawaban: C
Soal Notasi Sigma No. 14
$\sum\limits_{i=1}^{\infty }{\frac{i}{4^i}}$ = ….
A. $\frac{1}{3}$
B. $\frac{4}{9}$
C. $\frac{5}{9}$
D. $\frac{9}{4}$
E. $\frac{9}{5}$
Penyelesaian: Lihat/Tutup $S=\sum\limits_{i=1}^{\infty }{\frac{i}{4^i}}$
$\begin{align}S &= 4\sum\limits_{i=1}^{\infty }{\frac{i}{4^{i+1}}} \\ \frac{1}{4}S &= \sum\limits_{i=1}^{\infty }{\frac{i}{4^{i+1}}} \\ &= \sum\limits_{i=1+1}^{\infty }{\frac{i-1}{{{4}^{(i-1)+1}}}} \\ \frac{S}{4} &= \sum\limits_{i=2}^{\infty }{\frac{i-1}{4^i}}\,....\,(2) \end{align}$
Kurangkan persamaan (1) dan (2) maka:
$\begin{align}S-\frac{1}{4}S &= \sum\limits_{i=1}^{\infty }{\frac{i}{4^i}}-\sum\limits_{i=2}^{\infty }{\frac{i-1}{4^i}} \\ &= \frac{1}{4}+\sum\limits_{i=2}^{\infty }{\frac{i}{4^i}}-\sum\limits_{i=2}^{\infty }{\frac{i-1}{4^i}} \\ &= \frac{1}{4}+\sum\limits_{i=2}^{\infty }{\frac{1}{4^i}} \\ &= \frac{1}{4}+\left( \frac{1}{16}+\frac{1}{64}+\frac{1}{256}+... \right) \\ &= \frac{1}{4}+\frac{\frac{1}{16}}{1-\frac{1}{4}} \\ &= \frac{1}{4}+\frac{1}{16}.\frac{4}{3} \\ &= \frac{1}{4}+\frac{1}{12} \\ \frac{3}{4}S &= \frac{1}{3} \\ S &= \frac{4}{9} \end{align}$
Jawaban: B
Soal Notasi Sigma No. 15
$\sum\limits_{k=3}^{50}{\frac{1}{\sqrt{2k}+\sqrt{2k-2}}}$ = ….
A. 8
B. 7
C. 6
D. 5
E. 4
Penyelesaian: Lihat/Tutup $\sum\limits_{k=3}^{50}{\frac{1}{\sqrt{2k}+\sqrt{2k-2}}}$
= $\sum\limits_{k=3}^{50}{\frac{1}{\sqrt{2k}+\sqrt{2k-2}}\times \frac{\sqrt{2k}-\sqrt{2k-2}}{\sqrt{2k}-\sqrt{2k-2}}}$
= $\sum\limits_{k=3}^{50}{\frac{\sqrt{2k}-\sqrt{2k-2}}{2k-(2k-2)}}$
= $\sum\limits_{k=3}^{50}{\frac{\sqrt{2k}-\sqrt{2k-2}}{2}}$
= $\sum\limits_{k=3}^{50}{\frac{\sqrt{2k-2}-\sqrt{2k}}{-2}}$
= $\frac{\sqrt{4}-\sqrt{6}}{-2}+\frac{\sqrt{6}-\sqrt{8}}{-2}+\frac{\sqrt{8}-\sqrt{10}}{-2}+...+\frac{\sqrt{96}-\sqrt{98}}{-2}+\frac{\sqrt{98}-\sqrt{100}}{-2}$
= $\frac{\sqrt{4}-\sqrt{6}+\sqrt{6}-\sqrt{8}+\sqrt{8}-\sqrt{10}+...+\sqrt{96}-\sqrt{98}+\sqrt{98}-\sqrt{100}}{-2}$
= $\frac{\sqrt{4}-\sqrt{100}}{-2}$
= $\frac{2-10}{-2}$
= 4
Jawaban: E
Soal Notasi Sigma No. 16
Akar-akar persamaan kuadrat $x^2\sum\limits_{i=1}^{2}{2}+x.\sum\limits_{i=1}^{5}{i}-\frac{2}{3}\sum\limits_{i=1}^{3}{i}=0$ adalah $x_1$ dan $x_2$ dengan $x_1 < x_2$. Nilai dari $x_1+8x_2$ = ….
A. $-2$
B. $-1$
C. 0
D. 2
E. 4
Penyelesaian: Lihat/Tutup $\sum\limits_{i=1}^{2}{2}=2.2=4$
$\sum\limits_{i=1}^{5}{i}=\frac{5(5+1)}{2}=15$
$\sum\limits_{i=1}^{3}{i}=\frac{3(3+1)}{2}=6$
$\begin{align}x^2\sum\limits_{i=1}^{2}{2}+x.\sum\limits_{i=1}^{5}{i}-\frac{2}{3}\sum\limits_{i=1}^{3}{i} &= 0 \\ x^2.4+x.15-\frac{2}{3}.6 &= 0 \\ 4x^2+15x-4 &= 0 \\ (4x-1)(x+4) &= 0 \end{align}$
$4x-1=0\to x=\frac{1}{4}$
$x+4=0\to x=-4$
$x_1 < x_2$ maka $x_1=-4$ dan $x_2=\frac{1}{4}$
$x_1+8x_2=-4+8.\frac{1}{4}=-2$
Jawaban: A
Soal Notasi Sigma No. 17
Hasil dari $\sum\limits_{p=1}^{5}{\left( \sin \frac{p\pi }{2}+\cos \frac{p\pi }{2} \right)}$ = …
A. $-1$
B. $-\frac{1}{2}$
C. 0
D. $\frac{1}{2}$
E. 1
Penyelesaian: Lihat/Tutup $\sum\limits_{p=1}^{5}{\left( \sin \frac{p\pi }{2}+\cos \frac{p\pi }{2} \right)}$
= $\left( \sin \frac{\pi }{2}+\cos \frac{\pi }{2} \right)$ + $\left( \sin \pi +\cos \pi \right)$ + $\left( \sin \frac{3\pi }{2}+\cos \frac{3\pi }{2} \right)$ + $\left( \sin 2\pi +\cos 2\pi \right)$ + $\left( \sin \frac{5\pi }{2}+\cos \frac{5\pi }{2} \right)$
= $\left( 1+0 \right)$ + $\left( 0-1 \right)$ + $\left( -1+0 \right)$ + $\left( 0+1 \right)$ + $\left( 1+0 \right)$
= 1
Jawaban: E
Soal Notasi Sigma No. 18
Rumus $\sum\limits_{j=1}^{n}{(j+2)(j-5)}$ = ….
A. $\frac{n(n^2-3n-34)}{2}$
B. $\frac{n(n^2-3n-34)}{3}$
C. $\frac{n(n^2-3n-34)}{4}$
D. $\frac{n(n^2-3n-34)}{5}$
E. $\frac{n(n^2-3n-34)}{6}$
Penyelesaian: Lihat/Tutup $\sum\limits_{j=1}^{n}{(j+2)(j-5)}$
= $\sum\limits_{j=1}^{n}{(j^2-5j+2j-10)}$
= $\sum\limits_{j=1}^{n}{(j^2-3j-10)}$
= $\sum\limits_{j=1}^{n}{j^2}-3\sum\limits_{j=1}^{n}{j}-\sum\limits_{j=1}^{n}{10}$
= $\frac{1}{6}n(n+1)(2n+1)$$-3.\frac{1}{2}n(n+1)$$-10n$
= $\frac{(n^2+n)(2n+1)}{6}$$-\frac{3(n^2+n)}{2}$$-10n$
= $\frac{2n^3+3n^2+n}{6}$$-\frac{3n^2+3n}{2}$$-10n$
= $\frac{2n^3+3n^2+n}{6}-\frac{9n^2+9n}{6}-\frac{60n}{6}$
= $\frac{2n^3-6n^2-68n}{6}$
= $\frac{n^3-3n^2-34n}{3}$
= $\frac{n(n^2-3n-34)}{3}$
Jawaban: B
Soal Notasi Sigma No. 19
Nilai dari $\sum\limits_{n=5}^{10}{(2n-1)}$ = ….
A. 110
B. 94
C. 84
D. 76
E. 74
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{n=5}^{10}{(2n-1)} &= \sum\limits_{n=5-4}^{10-4}{(2(n+4)-1)} \\ &= \sum\limits_{n=1}^{6}{(2n+7)} \\ &= 2\sum\limits_{n=1}^{6}{n}+\sum\limits_{n=1}^{6}{7} \\ &= 2.\frac{6(6+1)}{2}+6.7 \\ &= 84 \end{align}$
Jawaban: C
Soal Notasi Sigma No. 20
$\sum\limits_{n=1}^{\infty }{\frac{2n+1}{5^n}}$ = ….
A. $\frac{7}{8}$
B. $\frac{6}{8}$
C. $\frac{5}{8}$
D. $\frac{4}{8}$
E. $\frac{3}{8}$
Penyelesaian: Lihat/Tutup $S=\sum\limits_{n=1}^{\infty }{\frac{2n+1}{5^n}}\,....\,(1)$
$\begin{align}S &= 5\sum\limits_{n=1}^{\infty }{\frac{2n+1}{5^{n+1}}} \\ \frac{1}{5}S &= \sum\limits_{n=1}^{\infty }{\frac{2n+1}{5^{n+1}}} \\ &= \sum\limits_{n=1+1}^{\infty +1}{\frac{2(n-1)+1}{{{5}^{(n-1)+1}}}} \\ \frac{1}{5}S &= \sum\limits_{n=2}^{\infty }{\frac{2n-1}{5^n}}\,....\,(2) \end{align}$
Persamaan (1) dikurangkan dengan persamaan (2) maka:
$\begin{align}S-\frac{1}{5}S &= \sum\limits_{n=1}^{\infty }{\frac{2n+1}{5^n}}-\sum\limits_{n=2}^{\infty }{\frac{2n-1}{5^n}} \\ \frac{4}{5}S &= \frac{3}{5}+\sum\limits_{n=2}^{\infty }{\frac{2n+1}{5^n}}-\sum\limits_{n=2}^{\infty }{\frac{2n-1}{5^n}} \\ &= \frac{3}{5}+\sum\limits_{n=2}^{\infty }{\frac{2}{5^n}} \\ &= \frac{3}{5}+2\sum\limits_{n=2}^{\infty }{\frac{1}{5^n}} \\ &= \frac{3}{5}+2\left( \frac{1}{25}+\frac{1}{125}+\frac{1}{625}+... \right) \\ &= \frac{3}{5}+2.\frac{\frac{1}{25}}{1-\frac{1}{5}} \\ &= \frac{3}{5}+2.\frac{1}{25}.\frac{5}{4} \\ &= \frac{3}{5}+\frac{1}{10} \\ \frac{4}{5}S &= \frac{7}{10} \\ S &= \frac{7}{8} \end{align}$
Jawaban: A
Soal Notasi Sigma No. 21
Notasi sigma yang menyatakan penjumlahan seribu bilangan asli pertama adalah ….
A. $\sum\limits_{n=1}^{1.000}{n}$
B. $\sum\limits_{n=1}^{1.000}{i}$
C. $\sum\limits_{n=1}^{1.000}{1}$
D. $\sum\limits_{n=1}^{999}{n}$
E. $\sum\limits_{n=1}^{999}{1}$
Penyelesaian: Lihat/Tutup $1+2+3+...+1000=\sum\limits_{n=1}^{1000}{n}$
Jawaban: A
Soal Notasi Sigma No. 22
Notasi sigma yang menyetakan penjumlahan 16 bilangan asli genap pertama adalah ….
A. $\sum\limits_{n=1}^{8}{2}$
B. $\sum\limits_{n=1}^{8}{2n}$
C. $\sum\limits_{n=1}^{16}{2}$
D. $\sum\limits_{n=1}^{16}{n}$
E. $\sum\limits_{n=1}^{16}{2n}$
Penyelesaian: Lihat/Tutup $2.1+2.2+2.3+...+2.16=\sum\limits_{n=1}^{16}{2n}$
Jawaban: E
Soal Notasi Sigma No. 23
Notasi sigma yang menyatakan penjumlahan beruntun 1 + 4 + 7 + 10 + 13 + 16 + 19 + 22 adalah ….
A. $\sum\limits_{n=1}^{8}{(4n-3)}$
B. $\sum\limits_{n=1}^{8}{(3n-2)}$
C. $\sum\limits_{n=2}^{9}{(4n-3)}$
D. $\sum\limits_{n=2}^{9}{(3n-2)}$
E. $\sum\limits_{n=0}^{7}{(3n+3)}$
Penyelesaian: Lihat/Tutup 1 + 4 + 7 + 10 + 13 + 16 + 19 + 22
= (3.1 – 2) + (3.2 – 2) + (3.3 – 2) + … + (3.8 – 2)
= $\sum\limits_{n=1}^{8}{(3n-2)}$
Jawaban: B
Soal Notasi Sigma No. 24
Nilai dari $\sum\limits_{i=1}^{10}{(i+2)}$ = ….
A. 63
B. 65
C. 73
D. 75
E. 77
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{i=1}^{10}{(i+2)} &= \sum\limits_{i=1}^{10}{i}+\sum\limits_{i=1}^{10}{2} \\ &= \frac{10(10+1)}{2}+10\times 2 \\ &= 55+20 \\ &= 75 \end{align}$
Jawaban: D
Soal Notasi Sigma No. 25
Nilai dari $\sum\limits_{i=3}^{8}{(4i-6)}$ = ….
A. 108
B. 102
C. 96
D. 88
E. 72
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{i=3}^{8}{(4i-6)} &= \sum\limits_{i=3-2}^{8-2}{(4(i+2)-6)} \\ &= \sum\limits_{i=1}^{6}{(4i+8-6)} \\ &= \sum\limits_{i=1}^{6}{(4i+2)} \\ &= 4\sum\limits_{i=1}^{6}{i}+\sum\limits_{i=1}^{6}{2} \\ &= 4.\frac{6(6+1)}{2}+6\times 2 \\ &= 96 \end{align}$
Jawaban: C
Soal Notasi Sigma No. 26
Nilai dari $\sum\limits_{k=1}^{5}{\frac{k-2}{k}}$ adalah ….
A. $\frac{1}{2}$
B. $\frac{13}{30}$
C. $\frac{2}{5}$
D. $\frac{11}{30}$
E. $\frac{1}{3}$
Penyelesaian: Lihat/Tutup $\sum\limits_{k=1}^{5}{\frac{k-2}{k}}$
= $\frac{1-2}{1}+\frac{2-2}{2}+\frac{3-2}{3}+\frac{4-2}{4}+\frac{5-2}{5}$
= $-1+0+\frac{1}{3}+\frac{1}{2}+\frac{3}{5}$
= $\frac{-1\times 30}{30}+\frac{1\times 10}{3\times 10}+\frac{1\times 15}{2\times 15}+\frac{3\times 6}{5\times 6}$
= $\frac{-30+10+15+18}{30}$
= $\frac{13}{30}$
Jawaban: B
Soal Notasi Sigma No. 27
Hasil dari $\sum\limits_{n=1}^{7}{3n}+\sum\limits_{n=1}^{7}{(n-1)}$ = ….
A. 84
B. 95
C. 102
D. 104
E. 105
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{n=1}^{7}{3n}+\sum\limits_{n=1}^{7}{(n-1)} &= \sum\limits_{n=1}^{7}{(3n+n-1)} \\ &= \sum\limits_{n=1}^{7}{(4n-1)} \\ &= 4\sum\limits_{n=1}^{7}{n}-\sum\limits_{n=1}^{7}{1} \\ &= 4.\frac{7(7+1)}{2}-7.1 \\ &= 105 \end{align}$
Jawaban: E
Soal Notasi Sigma No. 28
ilai dari $\sum\limits_{i=1}^{9}{(3i-2)\frac{i}{6}}$ adalah ….
A. 132,0
B. 131,5
C. 130,0
D. 127,5
E. 126,0
Penyelesaian: Lihat/Tutup $\sum\limits_{i=1}^{9}{(3i-2)\frac{i}{6}}$
= $\sum\limits_{i=1}^{9}{\left( \frac{1}{2}{{i}^{2}}-\frac{1}{3}i \right)}$
= $\frac{1}{2}\sum\limits_{i=1}^{9}{{{i}^{2}}}-\frac{1}{3}\sum\limits_{i=1}^{9}{i}$
= $\frac{1}{2}.\frac{9(9+1)(9.2+1)}{6}-\frac{1}{3}.\frac{9(9+1)}{2}$
= 142,5 – 15
= 127,5
Jawaban: D
Soal Notasi Sigma No. 29
Diketahui $\sum\limits_{k=1}^{10}{(2+ak)}=0$, nilai $a$ = ….
A. $-\frac{11}{4}$
B. $-\frac{4}{11}$
C. $-\frac{2}{11}$
D. $\frac{4}{11}$
E. $\frac{11}{2}$
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{k=1}^{10}{(2+ak)} &= 0 \\ \sum\limits_{k=1}^{10}{2}+a\sum\limits_{k=1}^{10}{k} &= 0 \\ 10.2+a.\frac{10(10+1)}{2} &= 0 \\ 20+55a &= 0 \\ 55a &= -20 \\ a &= -\frac{20}{55} \\ a &= -\frac{4}{11} \end{align}$
Jawaban: B
Soal Notasi Sigma No. 30
Diketahui $\sum\limits_{k=1}^{n}{(2k-1)}=225$, nilai $n$ = ….
A. 10
B. 15
C. 20
D. 15
E. 30
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{k=1}^{n}{(2k-1)} &= 225 \\ 2\sum\limits_{k=1}^{n}{k}-\sum\limits_{k=1}^{n}{1} &= 225 \\ 2.\frac{n(n+1)}{2}-n.1 &= 225 \\ n(n+1)-n &= 225 \\ n^2+n-n &= 225 \\ n^2 &= 225 \\ n &= 15 \end{align}$
Jawaban: B
Soal Notasi Sigma No. 31
Jika batas bawah diubah menjadi sepuluh, bentuk notasi sigma $\sum\limits_{n=2}^{12}{(n^2-3n)}$ akan menjadi ….
A. $\sum\limits_{n=10}^{20}{(n^2-19n+40)}$
B. $\sum\limits_{n=10}^{20}{(n^2-19n+88)}$
C. $\sum\limits_{n=10}^{20}{(n^2-13n+40)}$
D. $\sum\limits_{n=10}^{20}{(n^2-13n+88)}$
E. $\sum\limits_{n=10}^{20}{(n^2+13n+40)}$
Penyelesaian: Lihat/Tutup $\sum\limits_{n=2}^{12}{(n^2-3n)}$
= $\sum\limits_{n=2+8}^{12+8}{({{(n-8)}^{2}}-3(n-8))}$
= $\sum\limits_{n=10}^{20}{(n^2-16n+64-3n+24)}$
= $\sum\limits_{n=10}^{20}{(n^2-19n+88)}$
Jawaban: B
Soal Notasi Sigma No. 32
Jika bentuk $\sum\limits_{n=4}^{8}{{{(n-2)}^{2}}}$ diubah dengan notasi sigma lain yang sesuai dengan batas bawah satu, akan diperoleh bentuk ….
A. $\sum\limits_{n=1}^{5}{(n^2-2n+1)}$
B. $\sum\limits_{n=1}^{5}{(n^2+2n+1)}$
C. $\sum\limits_{n=1}^{5}{(n^2-4n-4)}$
D. $\sum\limits_{n=1}^{5}{(n^2-10n+25)}$
E. $\sum\limits_{n=1}^{5}{(n^2+10n+25)}$
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{n=4}^{8}{{{(n-2)}^{2}}} &= \sum\limits_{n=4-3}^{8-3}{{{((n+3)-2)}^{2}}} \\ &= \sum\limits_{n=1}^{5}{{{(n+1)}^{2}}} \\ &= \sum\limits_{n=1}^{5}{(n^2+2n+1)} \end{align}$
Jawaban: B
Soal Notasi Sigma No. 33
$\sum\limits_{i=1}^{n}{(4i-3)}$ = ….
A. $n(n+2)$
B. $n(2n-1)$
C. $n(2n+1)$
D. $n(3n-1)$
E. $n(3n+2)$
Penyelesaian: Lihat/Tutup $\sum\limits_{i=1}^{n}{(4i-3)}$
= $4\sum\limits_{i=1}^{n}{i}-\sum\limits_{i=1}^{n}{3}$
= $4.\frac{n(n+1)}{2}-n.3$
= $2n(n+1)-3n$
= $2n^2+2n-3n$
= $2n^2-n$
= $n(2n-1)$
Jawaban: B
Soal Notasi Sigma No. 34
$\sum\limits_{k=1}^{10}{(2^k-2^{k-1})}$ = ….
A. $1-{{2}^{10}}$
B. ${{2}^{10}}-1$
C. $1+{{2}^{10}}$
D. $2-{{2}^{9}}$
E. $2+{{2}^{10}}$
Penyelesaian: Lihat/Tutup $\sum\limits_{k=1}^{10}{(2^k-2^{k-1})}$
= $\sum\limits_{k=1}^{10}{(-2^{k-1}+2^k)}$
= $(-{{2}^{0}}+{{2}^{1}})+(-{{2}^{1}}+2^2)+(-2^2+{{2}^{3}})+...+(-{{2}^{9}}+{{2}^{10}})$
= $-{{2}^{0}}+{{2}^{1}}-{{2}^{1}}+2^2-2^2+{{2}^{3}}+...-{{2}^{9}}+{{2}^{10}}$
= $-{{2}^{0}}+{{2}^{10}}$
= $-1+{{2}^{10}}$
= ${{2}^{10}}-1$
Jawaban: B
Soal Notasi Sigma No. 35
Nilai dari $\sum\limits_{n=1}^{50}{[{{(-1)}^{n}}.n]}$ = ….
A. 50
B. 25
C. 0
D. $-25$
E. $-50$
Penyelesaian: Lihat/Tutup $\sum\limits_{n=1}^{50}{[{{(-1)}^{n}}.n]}$
= $-1+2-3+4-...-47+48-49+50$
= $\underbrace{1+1+1+...+1}_{sebanyak\,25}$
= 25
Jawaban: B
Soal Notasi Sigma No. 36
Nilai dari $\sum\limits_{n=1}^{160}{[{{(-1)}^{n}}(3n+7)]}$ = ….
A. 240
B. 80
C. 0
D. $-80$
E. $-240$
Penyelesaian: Lihat/Tutup $\sum\limits_{n=1}^{160}{[{{(-1)}^{n}}(3n+7)]}$
= $-10+13-16+19-....-478+481-484+487$
= $\underbrace{3+3+3+...+3+3}_{sebanyak\,80}$
= $80\times 3$
= 240
Jawaban: A
Soal Notasi Sigma No. 37
Bentuk notasi sigma yang bernilai sama dengan $\sum\limits_{k=1}^{n}{{{(k-3)}^{2}}}$ adalah ….
A. $\sum\limits_{k=1}^{n}{{{k}^{2}}}-9$
B. $\sum\limits_{k=1}^{n}{{{k}^{2}}}+6\sum\limits_{k=1}^{n}{k}+9$
C. $\sum\limits_{k=1}^{n}{{{k}^{2}}}-6\sum\limits_{k=1}^{n}{k}+9$
D. $\sum\limits_{k=1}^{n}{{{k}^{2}}}+6\sum\limits_{k=1}^{n}{k}+9n$
E. $\sum\limits_{k=1}^{n}{{{k}^{2}}}-6\sum\limits_{k=1}^{n}{k}+9n$
Penyelesaian: Lihat/Tutup $\begin{align}\sum\limits_{k=1}^{n}{{{(k-3)}^{2}}} &= \sum\limits_{k=1}^{n}{({{k}^{2}}-6k+9)} \\ &= \sum\limits_{k=1}^{n}{{{k}^{2}}}-6\sum\limits_{k=1}^{n}{k}+\sum\limits_{k=1}^{n}{9} \\ &= \sum\limits_{k=1}^{n}{{{k}^{2}}}-6\sum\limits_{k=1}^{n}{k}+9n \end{align}$
Jawaban: E
Soal Notasi Sigma No. 38
Bentuk notasi sigma yang bernilai sama dengan $\sum\limits_{p=1}^{n}{({{p}^{2}}-p+1)(p+1)}$ adalah ….
A. $\sum\limits_{p=1}^{n}{{{p}^{3}}}+n$
B. $\sum\limits_{p=1}^{n}{{{p}^{3}}}-n$
C. $\sum\limits_{p=1}^{n}{{{p}^{3}}}+2\sum\limits_{p=1}^{n}{{{p}^{2}}}+2\sum\limits_{p=1}^{n}{p}+n$
D. $\sum\limits_{p=1}^{n}{{{p}^{3}}}-2\sum\limits_{p=1}^{n}{{{p}^{2}}}+2\sum\limits_{p=1}^{n}{p}-n$
E. $\sum\limits_{p=1}^{n}{{{p}^{3}}}-2\sum\limits_{p=1}^{n}{{{p}^{2}}}-2\sum\limits_{p=1}^{n}{p}-n$
Penyelesaian: Lihat/Tutup $\sum\limits_{p=1}^{n}{({{p}^{2}}-p+1)(p+1)}$
= $\sum\limits_{p=1}^{n}{({{p}^{3}}+{{p}^{2}}-{{p}^{2}}-p+p+1)}$
= $\sum\limits_{p=1}^{n}{({{p}^{3}}+1)}$
= $\sum\limits_{p=1}^{n}{{{p}^{3}}}+\sum\limits_{p=1}^{n}{1}$
= $\sum\limits_{p=1}^{n}{{{p}^{3}}}+n$
Jawaban: A
Soal Notasi Sigma No. 39
Bentuk notasi sigma yang bernilai sama dengan $\sum\limits_{p=4}^{18}{{{(3p-1)}^{2}}}$ adalah ….
A. $9\sum\limits_{p=1}^{15}{{{p}^{2}}}+48\sum\limits_{p=1}^{15}{p}+64$
B. $9\sum\limits_{p=1}^{15}{{{p}^{2}}}+48\sum\limits_{p=1}^{15}{p}+960$
C. $9\sum\limits_{p=1}^{15}{{{p}^{2}}}-8\sum\limits_{p=1}^{15}{p}+64$
D. $9\sum\limits_{p=1}^{15}{{{p}^{2}}}-8\sum\limits_{p=1}^{15}{p}+960$
E. $9\sum\limits_{p=1}^{15}{{{p}^{2}}}+6\sum\limits_{p=1}^{15}{p}+15$
Penyelesaian: Lihat/Tutup $\sum\limits_{p=4}^{18}{{{(3p-1)}^{2}}}$
= $\sum\limits_{p=4-3}^{18-3}{{{(3(p+3)-1)}^{2}}}$
= $\sum\limits_{p=1}^{15}{{{(3p+9-1)}^{2}}}$
= $\sum\limits_{p=1}^{15}{{{(3p+8)}^{2}}}$
= $\sum\limits_{p=1}^{15}{(9{{p}^{2}}+48p+64)}$
= $9\sum\limits_{p=1}^{15}{{{p}^{2}}}+48\sum\limits_{p=1}^{15}{p}+\sum\limits_{p=1}^{15}{64}$
= $9\sum\limits_{p=1}^{15}{{{p}^{2}}}+48\sum\limits_{p=1}^{15}{p}+15\times 64$
= $9\sum\limits_{p=1}^{15}{{{p}^{2}}}+48\sum\limits_{p=1}^{15}{p}+960$
Jawaban: B
Soal Notasi Sigma No. 40
Nilai dari $\sum\limits_{k=1}^{\infty }{\frac{{{k}^{2}}}{{{7}^{k-1}}}}$ = ….
A. $\frac{45}{27}$
B. $\frac{46}{27}$
C. $\frac{47}{27}$
D. $\frac{48}{27}$
E. $\frac{49}{27}$
Penyelesaian: Lihat/Tutup $S=\sum\limits_{k=1}^{\infty }{\frac{{{k}^{2}}}{{{7}^{k-1}}}}\,....\,(1)$
$\begin{align}S &= 7\sum\limits_{k=1}^{\infty }{\frac{{{k}^{2}}}{{{7}^{k}}}} \\ \frac{1}{7}S &= \sum\limits_{k=1}^{\infty }{\frac{{{k}^{2}}}{{{7}^{k}}}} \\ &= \sum\limits_{k=1+1}^{\infty }{\frac{{{(k-1)}^{2}}}{{{7}^{k-1}}}} \\ \frac{1}{7}S &= \sum\limits_{k=2}^{\infty }{\frac{{{k}^{2}}-2k+1}{{{7}^{k-1}}}}\,....\,(2) \end{align}$
Persamaan (1) dikurangkan dengan persamaan (2) maka:
$\begin{align}S-\frac{1}{7}S &= \sum\limits_{k=1}^{\infty }{\frac{{{k}^{2}}}{{{7}^{k-1}}}}-\sum\limits_{k=2}^{\infty }{\frac{{{k}^{2}}-2k+1}{{{7}^{k-1}}}} \\ \frac{6}{7}S &= \frac{1}{{{7}^{1-1}}}+\sum\limits_{k=2}^{\infty }{\frac{{{k}^{2}}}{{{7}^{k-1}}}}-\sum\limits_{k=2}^{\infty }{\frac{{{k}^{2}}-2k+1}{{{7}^{k-1}}}} \\ \frac{6}{7}S &= 1+\sum\limits_{k=2}^{\infty }{\frac{2k-1}{{{7}^{k-1}}}}\,....\,(3) \end{align}$
$P=\sum\limits_{k=2}^{\infty }{\frac{2k-1}{{{7}^{k-1}}}}\,...\,(4)$
$\begin{align}P &= 7\sum\limits_{k=2}^{\infty }{\frac{2k-1}{{{7}^{k}}}} \\ \frac{1}{7}P &= \sum\limits_{k=2}^{\infty }{\frac{2k-1}{{{7}^{k}}}} \\ &= \sum\limits_{k=2+1}^{\infty +1}{\frac{2(k-1)-1}{{{7}^{k-1}}}} \\ \frac{1}{7}P &= \sum\limits_{k=3}^{\infty }{\frac{2k-3}{{{7}^{k-1}}}}\,....\,(5) \end{align}$
Persamaan (4) dikurangkan dengan persamaan (5) maka:
$\begin{align}P-\frac{1}{7}P &= \sum\limits_{k=2}^{\infty }{\frac{2k-1}{{{7}^{k-1}}}}\,-\sum\limits_{k=3}^{\infty }{\frac{2k-3}{{{7}^{k-1}}}} \\ \frac{6}{7}P &= \frac{2.2-1}{{{7}^{2-1}}}+\sum\limits_{k=3}^{\infty }{\frac{2k-1}{{{7}^{k-1}}}}-\sum\limits_{k=3}^{\infty }{\frac{2k-3}{{{7}^{k-1}}}} \\ &= \frac{3}{7}+\sum\limits_{k=3}^{\infty }{\frac{2}{{{7}^{k-1}}}} \\ &= \frac{3}{7}+\left( \frac{2}{{{7}^{2}}}+\frac{2}{{{7}^{3}}}+\frac{2}{{{7}^{4}}}+.... \right) \\ &= \frac{3}{7}+\frac{\frac{2}{49}}{1-\frac{1}{7}} \\ &= \frac{3}{7}+\frac{2}{49}.\frac{7}{6} \\ &= \frac{3}{7}+\frac{1}{21} \\ \frac{6}{7}P &= \frac{10}{21} \\ P &= \frac{10}{21}\times \frac{7}{6} \\ P &= \frac{5}{3}\times \frac{1}{3} \\ \sum\limits_{k=2}^{\infty }{\frac{2k-1}{{{7}^{k-1}}}} &= \frac{5}{9} \end{align}$
Substitusi ke persamaan (3):
$\begin{align}\frac{6}{7}S &= 1+\sum\limits_{k=2}^{\infty }{\frac{2k-1}{{{7}^{k-1}}}} \\ \frac{6}{7}S &= 1+\frac{5}{9} \\ \frac{6}{7}S &= \frac{14}{9} \\ S &= \frac{14}{9}\times \frac{7}{6} \\ S &= \frac{49}{27} \end{align}$
Jawaban: E
Soal Notasi Sigma No. 41
$\sum\limits_{i=1}^{15}{\frac{4}{(2n+3)(2n+1)}}$ = ….
A. $\frac{17}{33}$
B. $\frac{19}{33}$
C. $\frac{20}{33}$
D. $\frac{23}{33}$
E. $\frac{25}{33}$
Penyelesaian: Lihat/Tutup $\frac{4}{(2n+3)(2n+1)}=\frac{a}{2n+3}+\frac{b}{2n+1}$
$\frac{4}{(2n+3)(2n+1)}=\frac{a(2n+1)+b(2n+3)}{(2n+3)(2n+1)}$
$4=a(2n+1)+b(2n+3)$
$n=-\frac{1}{2}$ maka:
$4=a(2n+1)+b(2n+3)$
$4=a\left( 2\left( -\frac{1}{2} \right)+1 \right)+b\left( 2\left( -\frac{1}{2} \right)+3 \right)$
$4=a\left( -1+1 \right)+b\left( -1+3 \right)$
$4=2b$
$b=2$
$n=-\frac{3}{2}$ maka:
$4=a(2n+1)+b(2n+3)$
$4=a\left( 2\left( -\frac{3}{2} \right)+1 \right)+b\left( 2\left( -\frac{3}{2} \right)+3 \right)$
$4=a\left( -3+1 \right)+b\left( -3+3 \right)$
$4=-2a$
$a=-2$
Jadi,
$\sum\limits_{i=1}^{15}{\frac{4}{(2n+3)(2n+1)}}$
= $\sum\limits_{i=1}^{15}{\left[ \frac{a}{(2n+3)}+\frac{b}{(2n+1)} \right]}$
= $\sum\limits_{i=1}^{15}{\left[ \frac{-2}{(2n+3)}+\frac{2}{(2n+1)} \right]}$
= $\sum\limits_{i=1}^{15}{\left[ \frac{2}{(2n+1)}-\frac{2}{(2n+3)} \right]}$
= $\left( \frac{2}{3}-\frac{2}{5} \right)$ + $\left( \frac{2}{5}-\frac{2}{7} \right)$ + $\left( \frac{2}{7}-\frac{2}{9} \right)$ + … + $\left( \frac{2}{29}-\frac{2}{31} \right)$ + $\left( \frac{2}{31}-\frac{2}{33} \right)$
= $\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}$ + … + $\frac{2}{29}-\frac{2}{31}+\frac{2}{31}-\frac{2}{33}$
= $\frac{2}{3}-\frac{2}{33}$
= $\frac{22-2}{33}$
= $\frac{20}{33}$
Jawaban: C
Soal Notasi Sigma No. 42
$\sum\limits_{k=1}^{\infty }{\left( \frac{1}{(3k+1)(3k+4)} \right)}$ = ….
A. $\frac{1}{13}$
B. $\frac{1}{12}$
C. $\frac{1}{11}$
D. $\frac{1}{10}$
E. $\frac{1}{9}$
Penyelesaian: Lihat/Tutup $\begin{align}\frac{1}{(3k+1)(3k+4)} &= \frac{a}{3k+1}+\frac{b}{3k+4} \\ \frac{1}{(3k+1)(3k+4)} &= \frac{a(3k+4)+b(3k+1)}{(3k+1)(3k+4)} \\ 1 &= a(3k+4)+b(3k+1) \end{align}$
$k=-\frac{1}{3}$ maka:
$\begin{align}1 &= a(3k+4)+b(3k+1) \\ 1 &= 3a \\ a &= \frac{1}{3} \end{align}$
$k=-\frac{4}{3}$ maka:
$\begin{align}1 &= a(3k+4)+b(3k+1) \\ 1 &= -3b \\ b &= -\frac{1}{3} \end{align}$
$\sum\limits_{k=1}^{\infty }{\left( \frac{1}{(3k+1)(3k+4)} \right)}$
= $\sum\limits_{k=1}^{\infty }{\left( \frac{1/3}{3k+1}-\frac{1/3}{3k+4} \right)}$
= $\frac{1}{3}\sum\limits_{k=1}^{\infty }{\left( \frac{1}{3k+1}-\frac{1}{3k+4} \right)}$
= $\frac{1}{3}\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\left( \frac{1}{3k+1}-\frac{1}{3k+4} \right)}$
= $\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{3}\left( \frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{3n+1}-\frac{1}{3n+4} \right)$
= $\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{3}\left( \frac{1}{4}-\frac{1}{3n+4} \right)$
= $\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{3}\left( \frac{3n+4-4}{4(3n+4)} \right)$
= $\underset{n\to \infty }{\mathop{\lim }}\,\frac{n}{12n+16}$
= $\underset{n\to \infty }{\mathop{\lim }}\,\frac{\frac{n}{n}}{\frac{12n}{n}+\frac{16}{n}}$
= $\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{12+\frac{16}{n}}$
= $\frac{1}{12+\frac{16}{\infty }}$
= $\frac{1}{12+0}$
= $\frac{1}{12}$
Jawaban: B
Soal Notasi Sigma No. 43
Bentuk aljabar $x_1{{y}^{8}}+x_2{{y}^{9}}+{{x}_{3}}{{y}^{10}}+...+{{x}_{16}}{{y}^{23}}$ dapat dinyatakan dalam bentuk notasi sigma, yaitu ….
A. $\sum\limits_{n=0}^{16}{{{x}_{n}}.{{y}^{n+7}}}$
B. $\sum\limits_{n=1}^{8}{{{x}_{2n-1}}.{{y}^{2n+5}}}$
C. $\sum\limits_{n=3}^{18}{{{x}_{n-2}}.{{y}^{n+5}}}$
D. $\sum\limits_{n=2}^{17}{{{x}_{n-1}}.{{y}^{n+5}}}$
E. $\sum\limits_{n=-3}^{13}{{{x}_{n+3}}.{{y}^{n+8}}}$
Penyelesaian: Lihat/Tutup $x_1{{y}^{8}}+x_2{{y}^{9}}+{{x}_{3}}{{y}^{10}}+...+{{x}_{16}}{{y}^{23}}$
= $x_1{{y}^{1+7}}+x_2{{y}^{2+7}}+{{x}_{3}}{{y}^{3+7}}+...+{{x}_{16}}{{y}^{16+7}}$
= $\sum\limits_{n=1}^{16}{{{x}_{n}}{{y}^{n+7}}}$
= $\sum\limits_{n=1+2}^{16+2}{{{x}_{(n-2)}}{{y}^{(n-2)+7}}}$
= $\sum\limits_{n=3}^{18}{{{x}_{n-2}}.{{y}^{n+5}}}$
Jawaban: C
Soal Notasi Sigma No. 44
Nilai dari $\sum\limits_{n=1}^{70}{[{{(-1)}^{n}}(2n-5)]}$ = ….
A. 70
B. 35
C. 0
D. $-35$
E. $-70$
Penyelesaian: Lihat/Tutup $\sum\limits_{n=1}^{70}{[{{(-1)}^{n}}(2n-5)]}$
= $-(-3)+(-1)-1+3-...-129+131-133+135$
= $\underbrace{2+2+...+2+2}_{sebanyak\,35}$
= $35\times 2$
= 70
Jawaban: A
Soal Notasi Sigma No. 45
Nilai dari $\sum\limits_{i=1}^{20}{\left( \frac{1}{i}-\frac{1}{i+1} \right)}$ adalah ….
A. $\frac{1}{21}$
B. $\frac{1}{20}$
C. $\frac{9}{21}$
D. $\frac{19}{20}$
E. $\frac{20}{21}$
Penyelesaian: Lihat/Tutup $\sum\limits_{i=1}^{20}{\left( \frac{1}{i}-\frac{1}{i+1} \right)}$
= $\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}+\frac{1}{20}-\frac{1}{21}$
= $1-\frac{1}{21}$
= $\frac{20}{21}$
Jawaban: E
Soal Notasi Sigma No. 46
Nilai dari $\sum\limits_{n=1}^{10}{({{2}^{n}}-{{2}^{n-1}})}$ adalah ….
A. 1.025
B. 1.024
C. 1.023
D. $-1.023$
E. $-1.025$
Penyelesaian: Lihat/Tutup $\sum\limits_{n=1}^{10}{({{2}^{n}}-{{2}^{n-1}})}$
= $\sum\limits_{n=1}^{10}{(-{{2}^{n-1}}+{{2}^{n}})}$
= $-{{2}^{0}}+{{2}^{1}}-{{2}^{1}}+2^2-2^2+{{2}^{3}}-...-{{2}^{8}}+{{2}^{9}}-{{2}^{9}}+{{2}^{10}}$
= $-{{2}^{0}}+{{2}^{10}}$
= $-1+1024$
= 1023
Jawaban: C
Soal Notasi Sigma No. 47
Nilai dari $\sum\limits_{i=1}^{38}{\left( \frac{i}{i+1}-\frac{i+1}{i+2} \right)}$ adalah ….
A. $1\frac{19}{40}$
B. $\frac{19}{40}$
C. 0
D. $-\frac{19}{40}$
E. $-1\frac{19}{40}$
Penyelesaian: Lihat/Tutup $\sum\limits_{i=1}^{38}{\left( \frac{i}{i+1}-\frac{i+1}{i+2} \right)}$
= $\frac{1}{2}-\frac{2}{3}+\frac{2}{3}-\frac{3}{4}+\frac{3}{4}-\frac{4}{5}+...+\frac{37}{38}-\frac{38}{39}+\frac{38}{39}-\frac{39}{40}$
= $\frac{1}{2}-\frac{39}{40}$
= $-\frac{19}{40}$
Jawaban: D
Soal Notasi Sigma No. 48
Bentuk penjumlahan beruntun dari notasi sigma $\sum\limits_{n=1}^{4}{\left( \frac{1}{n+1}+\frac{1}{n} \right)}$ adalah ….
A. $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}$
B. $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}$
C. $\frac{1}{2}+\frac{5}{6}+\frac{1}{4}+\frac{9}{20}$
D. $\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{5}{6}$
E. $\frac{3}{2}+\frac{5}{6}+\frac{7}{12}+\frac{9}{20}$
Penyelesaian: Lihat/Tutup $\sum\limits_{n=1}^{4}{\left( \frac{1}{n+1}+\frac{1}{n} \right)}$
= $\sum\limits_{n=1}^{4}{\left( \frac{n}{n(n+1)}+\frac{n+1}{n(n+1)} \right)}$
= $\sum\limits_{n=1}^{4}{\left( \frac{2n+1}{n(n+1)} \right)}$
= $\frac{3}{2}+\frac{5}{6}+\frac{7}{12}+\frac{9}{20}$
Jawaban: E
Soal Notasi Sigma No. 49
Bentuk penjumlahan beruntun dari notasi sigma $\sum\limits_{n=1}^{5}{\left[ {{(-1)}^{n}}.\frac{n+1}{2n+1} \right]}$ adalah ….
A. $\frac{2}{3}+\frac{3}{5}+\frac{4}{7}+\frac{5}{9}+\frac{6}{11}$
B. $-\frac{2}{3}+\frac{3}{5}+\frac{4}{7}+\frac{5}{9}+\frac{6}{11}$
C. $-\frac{2}{3}+\frac{3}{5}-\frac{4}{7}+\frac{5}{9}-\frac{6}{11}$
D. $-\frac{2}{3}-\frac{3}{5}+\frac{4}{7}+\frac{5}{9}-\frac{6}{11}$
E. $-\frac{2}{3}-\frac{3}{5}-\frac{4}{7}-\frac{5}{9}-\frac{6}{11}$
Penyelesaian: Lihat/Tutup $\sum\limits_{n=1}^{5}{\left[ {{(-1)}^{n}}.\frac{n+1}{2n+1} \right]}=-\frac{2}{3}+\frac{3}{5}-\frac{4}{7}+\frac{5}{9}-\frac{6}{11}$
Jawaban: C
Soal Notasi Sigma No. 50
Notasi sigma yang menyatakan penjumlahan beruntun $2-4+8-16+32-64$ adalah ….
A. $-2\sum\limits_{n=1}^{6}{{{2}^{n-1}}}$
B. $-2\sum\limits_{n=1}^{6}{{{2}^{n+1}}}$
C. $-2\sum\limits_{n=1}^{6}{{{(-2)}^{n-1}}}$
D. $2\sum\limits_{n=1}^{6}{{{(-2)}^{n-1}}}$
E. $2\sum\limits_{n=1}^{6}{{{(-2)}^{n+1}}}$
Penyelesaian: Lihat/Tutup $2-4+8-16+32-64$
= ${{2.2}^{0}}-{{2.2}^{1}}+{{2.2}^{2}}-{{2.2}^{3}}+{{2.2}^{4}}-{{2.2}^{5}}$
= $2{{(-2)}^{0}}+2{{(-2)}^{1}}+2{{(-2)}^{2}}+2{{(-2)}^{3}}+2{{(-2)}^{4}}+2{{(-2)}^{5}}$
= $\sum\limits_{n=1}^{6}{2{{(-2)}^{n-1}}}$
= $2\sum\limits_{n=1}^{6}{{{(-2)}^{n-1}}}$
Jawaban: D

Post a Comment for "Kumpulan Soal - Notasi Sigma + Pembahasan"

Pertanyaan melalui kolom komentar akan direspon secepatnya. Jika tidak direspon, berarti pertanyaan serupa telah ada.