Question

If $S_n=\frac{3}{4}+\frac{5}{36}+\frac{7}{144}+\frac{9}{400}+...$ to $n$ terms, then find $\frac{1}{1-S_{40}}$.

Answer 1

$a=\frac{3}{4}$

$d=\frac{5}{36}-\frac{3}{4}=\frac{11}{18}$

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$S_{40}=\frac{40}{2}[2\left(\frac{3}{4}\right)+(40-1\left)\frac{11}{18}\right]$

$=20[\left(\frac{3}{2}\right)+(39\left)\frac{11}{18}\right]$

$=20\left[\frac{27+39\times 11}{18}\right]$

$=20\times \frac{76}{3}$

$=\frac{1520}{3}$

$\frac{1}{1-S_{40}}$

$=\frac{1}{1-1520}$

$=\frac{-1}{1519}$