Find the 20th term and the sum of 20 terms of the series

Solution:

Let $T_{n}$ be the $n$th term of the given series.

Thus, we have:

$T_{n}=2 n(2 n+2)=4 n^{2}+4 n$

For n = 20, we have:

$T_{20}=4 n^{2}+4 n$

$=4(20)^{2}+4(20)$

$=1600+80$

$=1680$

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

Now, let $S_{n}$ be the sum of $n$ terms of the given series.

Thus, we have:

$S_{n}=\sum_{k=1}^{n} T_{k}$

$\Rightarrow S_{n}=\sum_{k=1}^{n}\left(4 k^{2}+4 k\right)$

$\Rightarrow S_{n}=4 \sum_{k=1}^{n} k^{2}+4 \sum_{k=1}^{n} k$

For n = 20, we have

$S_{20}=4 \sum_{k=1}^{20} k^{2}+4 \sum_{k=1}^{20} k$

$\Rightarrow S_{20}=\frac{4(20)(21)(41)}{6}+\frac{4(20)(21)}{2}$

$\Rightarrow S_{20}=(40)(7)(41)+(40)(21)$

$\Rightarrow S_{20}=11480+840=12320$

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