# Thread: Sum of a series

1. ## Sum of a series

Given that $\displaystyle S_n$ is the sum of a geometric series where $\displaystyle S_n=1+\frac{a-2}{4}+\frac{(a-2)^2}{16}+...+\frac{(1-2)^{n-1}}{4^{n-1}}$

Find the values of $\displaystyle a$ such that $\displaystyle 256S_4=175S$

$\displaystyle 256S_4=175S$

$\displaystyle 256(1+\frac{a-2}{4}+\frac{(a-2)^2}{16}+\frac{(a-2)^3}{64})=175+175$

$\displaystyle (1+\frac{a-2}{4}+\frac{(a-2)^2}{16}+\frac{(a-2)^3}{64})=1\frac{47}{128}$

$\displaystyle \frac{a-2}{4}+\frac{(a-2)^2}{16}+\frac{(a-2)^3}{64})=\frac{47}{128}$

$\displaystyle 16(a-2)+4(a-2)^2+(a-2)^3=23.5$

2. Originally Posted by Punch
Given that $\displaystyle S_n$ is the sum of a geometric series where $\displaystyle S_n=1+\frac{a-2}{4}+\frac{(a-2)^2}{16}+...+\frac{(1-2)^{n-1}}{4^{n-1}}$

Find the values of $\displaystyle a$ such that $\displaystyle 256S_4=175S$

$\displaystyle 256S_4=175S$
what is S??

$\displaystyle 256(1+\frac{a-2}{4}+\frac{(a-2)^2}{16}+\frac{(a-2)^3}{64})=175+175$
how did you get 175+175 in the RHS??

$\displaystyle (1+\frac{a-2}{4}+\frac{(a-2)^2}{16}+\frac{(a-2)^3}{64})=1\frac{47}{128}$

$\displaystyle \frac{a-2}{4}+\frac{(a-2)^2}{16}+\frac{(a-2)^3}{64})=\frac{47}{128}$

$\displaystyle 16(a-2)+4(a-2)^2+(a-2)^3=23.5$
...

3. Originally Posted by Punch
Given that $\displaystyle S_n$ is the sum of a geometric series where $\displaystyle S_n=1+\frac{a-2}{4}+\frac{(a-2)^2}{16}+...+\frac{(1-2)^{n-1}}{4^{n-1}}$
I am going to assume that there should be "a- 2" in that last term.

Find the values of $\displaystyle a$ such that $\displaystyle 256S_4=175S$
I am going to assume that "S" is the sum of the infinite series, which is $\displaystyle \frac{4}{6-a}$.

$\displaystyle 256S_4=175S$

$\displaystyle 256(1+\frac{a-2}{4}+\frac{(a-2)^2}{16}+\frac{(a-2)^3}{64})=175+175$
How did 175S become 175+ 175? You should have
$\displaystyle 256\left(1+ \frac{a-2}{4}+ \frac{(a- 2)^2}{16}+ \frac{a-2)^3}{64}\right)= \frac{700}{6- a}= \frac{700}{4- (a-2)}$
if you let y= a- 2 that becomes
$\displaystyle 256\left(1+ \frac{y}{4}+ \frac{y^2}{16}+ \frac{y^3}{64}\right)= \frac{700}{4-y}$
$\displaystyle \left(1+ \frac{y}{4}+ \frac{y^2}{16}+ \frac{y^3}{64}\right)= \frac{175}{64(4- y)}$

Multiplying on both sides by 64:
$\displaystyle 64+ 16y+ 4y^2+ y^3= \frac{175}{4-y}$

Multiplying both sides by 4- y:
$\displaystyle 256- y^4= 175$
so the whole problem reduces to
$\displaystyle y^4= 81$

$\displaystyle (1+\frac{a-2}{4}+\frac{(a-2)^2}{16}+\frac{(a-2)^3}{64})=1\frac{47}{128}$

$\displaystyle \frac{a-2}{4}+\frac{(a-2)^2}{16}+\frac{(a-2)^3}{64})=\frac{47}{128}$

$\displaystyle 16(a-2)+4(a-2)^2+(a-2)^3=23.5$