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:

\[S_n = \sum^n_{k = 1} T_k \]

\[ = \sum^n_{k = 1} \left[ 2k - 1 \right]^3 \]

\[ = \sum^n_{k = 1} \left[ 8 k^3 - 1 - 6k\left( 2k - 1 \right) \right]\]

\[ = \sum^n_{k = 1} \left[ 8 k^3 - 1 - 12 k^2 + 6k \right]\]

\[ = \sum^n_{k = 1} \left[ 8 k^3 - 1 - 12 k^2 + 6k \right]\]

\[ = {8\sum}^n_{k = 1} k^3 - \sum^n_{k = 1} 1 - 12 \sum^n_{k = 1} k^2 + {6\sum}^n_{k = 1} k \]

\[ = \frac{8 n^2 \left( n + 1 \right)^2}{4} - n - \frac{12n\left( n + 1 \right)\left( 2n + 1 \right)}{6} + \frac{6 n\left( n + 1 \right)}{2}\]

\[ = 2 n^2 \left( n + 1 \right)^2 - n - 2n\left( n + 1 \right)\left( 2n + 1 \right) + 3n\left( n + 1 \right)\]

\[ = n\left( n + 1 \right)\left[ 2n\left( n + 1 \right) - 2\left( 2n + 1 \right) + 3 \right] - n\]

\[ = n\left( n + 1 \right)\left[ 2 n^2 - 2n + 1 \right] - n\]

\[ = n\left[ 2 n^3 - 2 n^2 + n + 2 n^2 - 2n + 1 - 1 \right]\]

\[ = n\left[ 2 n^3 - n \right]\]

\[ = n^2 \left[ 2 n^2 - 1 \right]\]