Question

Find the sum of the geometric series 512+256+ . . .+4

a. 1,022
b. 1,016
c. 510
d. 1,020

Answer 1

`\bf 512~~,~~\stackrel{512\cdot (1)/(2)}{256}~~,~~...4`

so, as you can see above, the common ratio r = 1/2, now, what term is +4 anyway?

`\bf n^(th)\textit{ term of a geometric sequence}\\\\a_n=a_1\cdot r^(n-1)\qquad \begin{cases}n=n^(th)\ term\\a_1=\textit{first term's value}\\r=\textit{common ratio}\\----------\\r=(1)/(2)\\a_1=512\\a_n=+4\end{cases}`

`\bf 4=512\left( \cfrac{1}{2} \right)^(n-1)\implies \cfrac{4}{512}=\left( \cfrac{1}{2} \right)^(n-1)\\\\\\\cfrac{1}{128}=\left( \cfrac{1}{2} \right)^(n-1)\implies \cfrac{1}{2^7}=\left( \cfrac{1}{2} \right)^(n-1)\implies 2^(-7)=\left( 2^(-1)\right)^(n-1)\\\\\\(2^(-1))^7=(2^(-1))^(n-1)\implies 7=n-1\implies \boxed{8=n}`

so is the 8th term, then, let's find the Sum of the first 8 terms.

`\bf \qquad \qquad \textit{sum of a finite geometric sequence}\\\\S_n=\sum\limits_(i=1)^(n)\ a_1\cdot r^(i-1)\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases}n=n^(th)\ term\\a_1=\textit{first term's value}\\r=\textit{common ratio}\\----------\\r=(1)/(2)\\a_1=512\\n=8\end{cases}`

`\bf S_8=512\left[ \cfrac{1-\left( (1)/(2) \right)^8}{1-(1)/(2)} \right]\implies S_8=512\left(\cfrac{1-(1)/(256)}{(1)/(2)} \right)\implies S_8=512\left(\cfrac{(255)/(256)}{(1)/(2)} \right)\\\\\\S_8=512\cdot \cfrac{255}{128}\implies S_8=1020`