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 $t_{n} = \frac{n^{2} \left(n + 1\right)^{2}}{4\cdot n^{2}} = \frac{\left(n +1\right)^{2}}{4} = \frac{1}{4} \left(n^{2} + 2n + 1\right)$
$\therefore S_{n} = \frac{1}{4} \left[\sum\limits^{n}_{k =1}k^{2} +2\sum\limits^{n}_{k = 1}k + n\right]$
$= \frac{1}{4}\left[\frac{n\left(n + 1\right) \left(2n + 1\right)}{6} +\frac{2n\left(n + 1\right)}{2} +n \right]$
$\therefore S_{16} = \frac{1}{4} \left[ \frac{16\cdot17\cdot33}{6} + 16\cdot 17 + 16\right]$
$= \frac{1}{4}\left[1496 + 272 + 16\right] = \frac{1}{4} \left[1784\right] =446$