Which term of the $GP.,$ $2,8,32, \ldots$ up to $n$ terms is $131072 ?$
Let $131072$ be the $n^{\text {th }}$ term of the given $G.P.$ Here $a=2$ and $r=4$
Therefore $\quad 131072=a_{n}=2(4)^{n-1} \quad$ or $\quad 65536=4^{n-1}$
This gives $4^{8}=4^{n-1}$
So that $n-1=8,$ i.e., $n=9 .$ Hence, $131072$ is the $9^{\text {th }}$ term of the $G.P.$
Let for $n =1,2, \ldots \ldots, 50, S _{ a }$ be the sum of the infinite geometric progression whose first term is $n ^{2}$ and whose common ratio is $\frac{1}{(n+1)^{2}}$. Then the value of $\frac{1}{26}+\sum\limits_{n=1}^{50}\left(S_{n}+\frac{2}{n+1}-n-1\right)$ is equal to
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