What is the sum of the first 10 terms given 1, 1/3, 1/9,1/27..?

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What is the sum of the first 10 terms given 1, 1/3, 1/9,1/27..?

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Answer 1 · 2018-07-07T17:51:38

$\color{red}{3/2(1-(1/3)^{10})\approx1.499974597$

Explanation:

Given series $1, 1/3, 1/9, 1/27, \ldots$ is a G.P. with the first term $a=1$ & common ratio $r=\frac{1/3}{1}=\frac{1/9}{1/3}=\frac{1/27}{1/9}=\ldots =1/3$

Now, the sum of given series up to $n=10$ terms

$=\frac{a(r^{n}-1)}{r-1}$

$=\frac{1((1/3)^{10}-1)}{1/3-1}$

$=\frac{(1/3)^{10}-1}{-2/3}$

$=3/2(1-(1/3)^{10})$

$approx1.499974597$

Answer 2 · 2018-07-07T18:10:23

$S_10=3/2((3^10-1)/3^10)~~1.4999$

Explanation:

Here,

$1,1/3,1/9,1/27,...,$

The first term $=a_1=1$

The common ratio $=r=(1/3)/1=(1/9)/(1/3)=(1/27)/(1/9)=1/3$

So, the given sequence is Geometric sequence with

$color(blue)(a_1=1 and r=1/3$

The sum of first $n-terms$ of geometric sequence is :

$color(blue)(S_n=(a_1(1-r^n) )/(1-r) )and$ we have $color(blue)(n=10$

So,

$S_10=(1(1-(1/3)^10))/(1-1/3)$

$S_10=((1-1/3^10))/(2/3)$

$S_10=3/2((3^10-1)/3^10)$

$S_10=3/2(59048/59049)$

$S_10~~1.4999$

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What is the sum of the first 10 terms given 1, 1/3, 1/9,1/27..?

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