Question

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

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

Answer 1

The correct option is A $446$

$S_n=1+\frac{9}{4}+\frac{36}{9}+\frac{100}{16}+\cdots$ upto $n$ terms
$\Rightarrow S_n=1^3+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+...$
Now, $t_n=\frac{1^3+2^3+3^3+....+n^3}{1+3+5+....+(2n-1)}=\frac{n^2{(n+1)}^2}{4n^2}=\frac{{(n+1)}^2}{4}$
$\Rightarrow t_n=\frac{n^2+2n+1}{4}$
$S_n={\sum }_{n=1}^n{t}_n={\sum }_{n=1}^n\left(\frac{n^2+2n+1}{4}\right)$
$\Rightarrow S_n=\frac{1}{4}{\sum }_{n=1}^n{n}^2+\frac{1}{2}{\sum }_{n=1}^{n}n+\frac{1}{4}{\sum }_{n=1}^{n}1$
$\Rightarrow S_n=\frac{n(n+1)(2n+1)}{24}+\frac{n(n+1)}{4}+n$
$\Rightarrow S_n=\frac{n(2n^2+9n+13)}{24}$
$\Rightarrow S_{16}=\frac{16({216}^2+9.16+13)}{24}=446$
Ans: A