Question

The sum of the series $1+\frac{9}{4}+\frac{36}{9}+\frac{100}{16}+...$ up to $n$ terms if $n=16$ is

• A. $446$
• B. $746$
• C. $546$
• D. $846$

Answer 1

Answer: (A)

The given series can be written as
$1^3+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+...$
$t_n=\frac{1^3+2^3+3^3+....+n^3}{1+3+5+...+\left(2n-1\right)}$
or $t_n=\frac{n^2{\left(n+1\right)}^2}{4\cdot n^2}=\frac{{\left(n+1\right)}^2}{4}=\frac{1}{4}\left(n^2+2n+1\right)$
$\therefore S_n=\frac{1}{4}\left[\sum _{k=1}^n{k}^2+2\sum _{k=1}^{n}k+n\right]$
$=\frac{1}{4}\left[\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{2n\left(n+1\right)}{2}+n\right]$
$\therefore S_{16}=\frac{1}{4}\left[\frac{16\cdot 17\cdot 33}{6}+16\cdot 17+16\right]$
$=\frac{1}{4}\left[1496+272+16\right]=\frac{1}{4}\left[1784\right]=446$