Respuesta :

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Let [tex]S_N[/tex] denote the [tex]N[/tex]th partial sum of the series. Then

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

Multiplying both sides by [tex]\dfrac12[/tex] gives

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

Subtracting the last expression from [tex]S_N[/tex], we have

[tex]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)[/tex]
[tex]\displaystyle\frac12S_N=-144\left(1-\frac1{2^{N+1}}\right)[/tex]
[tex]\displaystyle S_N=-288\left(1-\frac1{2^{N+1}}\right)[/tex]

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

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

so the answer is A.

Answer:

Option A is correct that is -288

Explanation:

Formula for sum of infinite geometric progression  [tex]S_{\infty} =\frac{a}{1-r}[/tex]

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

[tex]\Rightarrow \frac{-144}{1-\frac{1}{2}}\\\\\Rightarrow -288[/tex]

sum of the infinite geometric series= -288

So, Option A is correct that is -288


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