Question

1³ + 3³ + 5³ + 7³ + ...

Answer 1

Let $T_n$ be the nth term of the given series.

Thus, we have:

$T_n={\left(2n-1\right)}^3$

Now, let $S_n$ be the sum of n terms of the given series.

Thus, we have:

$$\begin{gathered} S_n=\sum _{k=1}^n{T}_k \\ =\sum _{k=1}^n{\left[2k-1\right]}^3 \\ =\sum _{k=1}^n\left[8k^3-1-6k\left(2k-1\right)\right] \\ =\sum _{k=1}^n\left[8k^3-1-12k^2+6k\right] \\ =\sum _{k=1}^n\left[8k^3-1-12k^2+6k\right] \\ =\underset{k=1}{\overset{n}{8\sum }}{k}^3-\sum _{k=1}^{n}1-12\sum _{k=1}^n{k}^2+\underset{k=1}{\overset{n}{6\sum }}k \\ =\frac{8n^2{\left(n+1\right)}^2}{4}-n-\frac{12n\left(n+1\right)\left(2n+1\right)}{6}+\frac{6n\left(n+1\right)}{2} \\ =2n^2{\left(n+1\right)}^2-n-2n\left(n+1\right)\left(2n+1\right)+3n\left(n+1\right) \\ =n\left(n+1\right)\left[2n\left(n+1\right)-2\left(2n+1\right)+3\right]-n \\ =n\left(n+1\right)\left[2n^2-2n+1\right]-n \\ =n\left[2n^3-2n^2+n+2n^2-2n+1-1\right] \\ =n\left[2n^3-n\right] \\ =n^2\left[2n^2-1\right] \end{gathered}$$