Question

2² + 4² + 6² + 8² + ...

Answer 1

Let \[T_n\] be the nth term of the given series.

Thus, we have: \[T_n = \left( 2n \right)^2\]

Now, let \[S_n\] be the sum of n terms of the given series.

Thus, we have:

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

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

\[ = \sum^n_{k = 1} 4 k^2 \]

\[ = 4 \sum^n_{k = 1} k^2 \]

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

\[ = \frac{2n}{3}\left( n + 1 \right)\left( 2n + 1 \right)\]