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:

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

For n = 20, we have:

$$\begin{gathered} T_{20}=4n^2+4n \\ =4{\left(20\right)}^2+4\left(20\right) \\ =1600+80 \\ =1680 \end{gathered}$$

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:
$S_n=\sum _{k=1}^n{T}_k$

$$\begin{gathered} \Rightarrow S_n=\sum _{k=1}^n\left(4k^2+4k\right) \\ \Rightarrow S_n=\underset{k=1}{\overset{n}{4\sum }}{k}^2+4\sum _{k=1}^{n}k \end{gathered}$$

For n = 20, we have:

$$\begin{gathered} S_{20}=\underset{k=1}{\overset{20}{4\sum }}{k}^2+4\sum _{k=1}^{20}k \\ \Rightarrow S_{20}=\frac{4\left(20\right)\left(21\right)\left(41\right)}{6}+\frac{4\left(20\right)\left(21\right)}{2} \\ \Rightarrow S_{20}=\left(40\right)\left(7\right)\left(41\right)+\left(40\right)\left(21\right) \\ \Rightarrow S_{20}=11480+840=12320 \end{gathered}$$

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