Question

If $S=4-9x+16x^2-25x^3+36x^4-49x^5+......\infty$. Find the value of S for x=$\frac{-1}{2}$.

Answer 1

This series is not AP,GP,HP and AGP
But we see 4,9,16,25.......$\infty$ are square numbers and $1,(-x),(-x{)}^2,(-x{)}^3,(-x{)}^4$,.......$\infty$ is a geometric progression. if we subtract square numbers. We can form an AP.
So, multiply given series by common ratio and then subtract it with original series.
Let $s=4-9x+16x^2-25x^3+36x^4-49x^5+......\infty$ -----(1)
Multiply by common difference in equation 1
$-xs=-4x+9x^2-16x^3+25x^4-36x^5+......\infty$ -----(2)
) On subtracting equation 2 from equation 1
We get,
$s(1+x)=4-5x+7x^2-9x^3+11x^4-13x^5+......\infty$ -----(3)
Multiply by common difference in equation 3
$-s(1+x)x=-4x+5x^2-7x^3+9x^4-11x^5+......\infty$ -----(4)
On subtracting equation 4 from equation 3
$s(1+x{)}^2=4-x+2x^2-2x^3+2x^4-2x^5+......\infty$
$s(1+x{)}^2=4-x+2(x^2-x^3+x^4-x^5+......\infty )$
$s(1+x{)}^2=4-x+2x^2(1-x+x^2-......\infty )$
$s(1+x{)}^2=4-x+\frac{2x^2.1}{1+x}$
$s(1+x{)}^2=4-x+\frac{2x^2}{1+x}=\frac{4+3x+x^2}{x+1}$
$s=\frac{x^2+3x+4}{(x+1{)}^3}$
When x=-$\frac{1}{2}$
s=$\frac{(\frac{-1}{2}{)}^2+3(\frac{-1}{2})+4}{(\frac{-1}{2}+1{)}^3}$
$\frac{\frac{1}{4}-\frac{3}{2}+4}{\frac{1}{3}}=22$