Question

Sum of $n$ terms of the following series $1^3+3^3+5^3+7^3+............$ is

Answer 1

Finding the sum of $n$ terms of the following series

The given series is: $1^3+3^3+5^3+7^3+............$

Here, the $n^{th}$ term of the series can be given as

${\mathrm{T}}_n={\left(2n-1\right)}^3......\left(1\right)$

We know that

${\left(a-b\right)}^3=a^3-3a^{2}b+3a{b}^2-b^3$

Then, Equation $\left(1\right)$ becomes

$T_p={\left(2p-1\right)}^3$

$$\begin{gathered} {\mathrm{T}}_p={\left(2p\right)}^3-\left(3\times {\left(2p\right)}^2\times 1\right)+\left(3\times 2p\times {\left(1\right)}^2\right)-{\left(1\right)}^3 \\ =8p^3-12p^2+6p-1 \end{gathered}$$

Now, the sum ${\mathrm{S}}_{\mathrm{n}}$ $=$ Sum of $\text{'}n\text{'}$ terms $=\sum _{\mathrm{p}=1}^{\mathrm{n}}{\mathrm{T}}_{\mathrm{p}}$

Since,

$\begin{array}{rcl}{\mathrm{S}}_{\mathrm{n}}& =& \sum _{\mathrm{p}=1}^{\mathrm{n}}\left(8{\mathrm{p}}^3-12{\mathrm{p}}^2+6\mathrm{p}-1\right)\\ & =& 8\times \sum _{\mathrm{p}=1}^{\mathrm{n}}{\mathrm{p}}^3-12\times \sum _{\mathrm{p}=1}^{\mathrm{n}}{\mathrm{p}}^2+6\times \sum _{\mathrm{p}=1}^{\mathrm{n}}\mathrm{p}-\sum _{\mathrm{p}=1}^{\mathrm{n}}1\\ & =& 8\times \frac{{\mathrm{n}}^2{\left(\mathrm{n}+1\right)}^2}{4}-12\times \frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+6\times \frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}-\mathrm{n}\\ & =& 2{\mathrm{n}}^2{\left(\mathrm{n}+1\right)}^2-2\left({\mathrm{n}}^2+\mathrm{n}\right)\left(2\mathrm{n}+1\right)+3\left({\mathrm{n}}^2+\mathrm{n}\right)-\mathrm{n}\\ & =& 2{\mathrm{n}}^2\left({\mathrm{n}}^2+2\mathrm{n}+1\right)-2\left(2{\mathrm{n}}^3+3{\mathrm{n}}^2+\mathrm{n}\right)+3{\mathrm{n}}^2+3\mathrm{n}-\mathrm{n}\\ & =& 2{\mathrm{n}}^4+4{\mathrm{n}}^3+2{\mathrm{n}}^2-4{\mathrm{n}}^3-6{\mathrm{n}}^2-2\mathrm{n}+3{\mathrm{n}}^2+3\mathrm{n}-\mathrm{n}\\ & =& 2{\mathrm{n}}^4-{\mathrm{n}}^2\\ & =& {\mathrm{n}}^2\left(2{\mathrm{n}}^2-1\right)\end{array}$

Hence, the correct answer is ${\mathrm{n}}^2\left(2{\mathrm{n}}^2-1\right)$.