Question

What is the sum of the infinite geometric series?
A. –288
B. –216
C. –144
D. –72

Answer 1

Let $S_N$ denote the $N$th partial sum of the series. Then

$S_N=\displaystyle\sum_{n=1}^N(-144)\left(\frac12\right)^{n-1}$
$S_N=\displaystyle-144\left(1+\frac12+\frac1{2^2}+\cdots+\frac1{2^{N-1}}+\frac1{2^N}\right)$

Multiplying both sides by $\dfrac12$ gives

$\dfrac12S_N=\displaystyle\sum_{n=1}^N(-144)\left(\frac12\right)^n$
$\dfrac12S_N=\displaystyle-144\left(\frac12+\frac1{2^2}+\frac1{2^3}+\cdots+\frac1{2^N}+\frac1{2^{N+1}}\right)$

Subtracting the last expression from $S_N$, we have

$S_N-\frac12S_N=-144\left(1+\frac12+\dfrac1{2^2}+\cdots+\dfrac1{2^{N-1}}+\dfrac1{2^N}\right)-(-144)\left(\dfrac12+\dfrac1{2^2}+\dfrac1{2^3}+\cdots+\dfrac1{2^N}+\dfrac1{2^{N+1}}\right)$
$\displaystyle\frac12S_N=-144\left(1-\frac1{2^{N+1}}\right)$
$\displaystyle S_N=-288\left(1-\frac1{2^{N+1}}\right)$

As $N\to\infty$, we end up with the original infinite series. The rational term approaches 0, leaving us with

$\displaystyle\sum_{n=1}^\infty(-144)\left(\frac12\right)^{n-1}=\lim_{n\to\infty}S_N=-288$

so the answer is A.

Answer 2

Answer:

Option A is correct that is -288

Explanation:

Formula for sum of infinite geometric progression  $S_{\infty} =\frac{a}{1-r}$

where,

a is the first of geometric series

r is the common ratio of geometric series

here, a= -144 and r=1/2

Now, substituting the values in the formula we will get

$\Rightarrow \frac{-144}{1-\frac{1}{2}}\\\\\Rightarrow -288$

sum of the infinite geometric series= -288

So, Option A is correct that is -288