Sum to n terms of the series 131+13+231+3+13+23+331+3+5+…is

A
$\frac{n}{24} (n^{2} + 9 n + 13)$
B
$\frac{n}{24} (2 n^{2} + 7 n + 15)$
C
$\frac{n}{24} (2 n^{2} + 9 n + 13)$
D
$\frac{n}{24} (n^{2} + 11 n + 11)$

Solution:

Let $t_{r}$ denote the $r t h$ term of the series, then
$t_{r} = \frac{1^{3} + 2^{3} + \dots + r^{3}}{1 + 3 + \dots + (2 r - 1)} = \frac{\frac{1}{4} r^{2} (r + 1)^{2}}{r^{2}}$
$\begin{array}{l} = \frac{1}{4} (r + 1)^{2} \\ \Rightarrow \sum_{r = 1}^{n} t_{r} = \frac{1}{4} \sum_{r = 1}^{n} (r + 1)^{2} = \frac{1}{4} [\sum_{r = 1}^{n + 1} r^{2} - 1] \\ = \frac{1}{4} [\frac{(n + 1) (n + 2) (2 n + 3)}{6} - 1] \\ = \frac{1}{24} [2 n^{3} + 9 n^{2} + 13 n + 6 - 6] \\ = \frac{n}{24} (2 n^{2} + 9 n + 13) \end{array}$

Sum to n terms of the series $\frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{1 + 3} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 3 + 5} + \dots$ is

Solution:

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