Question

If sum of the series $1\cdot 1+3\cdot 01+5\cdot 001+7\cdot 0001$ upto $n$ terms be $n^2+\frac{1}{m}(1-\frac{1}{k^n})$, then find $k-m$.

Answer 1

$S=1\cdot 1+3\cdot 01+5\cdot 001+7\cdot 0001$ upto $n$ terms $=(1+3+5+...+2n-1)+({10}^{-1}+{10}^{-2}+{10}^{-3}+...+{10}^{-n})=$ sum of A.P + sum of G.P
$S=\frac{n(2+(n-1)2)}{2}+\frac{\frac{1}{10}(\frac{1}{{10}^n}-1)}{\frac{1}{10}-1}=n^2+\frac{1}{9}(1-\frac{1}{{10}^n})$
Therefore, $k-m=10-9=1$