Question

Find the 20^th term and the sum of 20 terms of the series 2 × 4 + 4 × 6 + 6 × 8 + ...

Answer 1

Let \[T_n\] be the nth term of the given series.
Thus, we have:

\[T_n = 2n\left( 2n + 2 \right) = 4 n^2 + 4n\]

For n = 20, we have:

\[T_{20} = 4 n^2 + 4n\]

\[ = 4 \left( 20 \right)^2 + 4\left( 20 \right)\]

\[ = 1600 + 80\]

\[ = 1680\]

Therefore, the 20th term of the given series is 1680.
Now, let

\[S_n\] be the sum of n terms of the given series.
Thus, we have:

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

\[\Rightarrow S_n = \sum^n_{k = 1} \left( 4 k^2 + 4k \right)\]

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

For n = 20, we have:

\[S_{20} = {4\sum}^{20}_{k = 1} k^2 + 4 \sum^{20}_{k = 1} k\]

\[ \Rightarrow S_{20} = \frac{4\left( 20 \right)\left( 21 \right)\left( 41 \right)}{6} + \frac{4\left( 20 \right)\left( 21 \right)}{2}\]

\[ \Rightarrow S_{20} = \left( 40 \right)\left( 7 \right)\left( 41 \right) + \left( 40 \right)\left( 21 \right)\]

\[ \Rightarrow S_{20} = 11480 + 840 = 12320\]

Hence, the sum of the first 20 terms of the series is 12320.