Question

Find the ${20}^{th}$ term and the sum of $20$ terms of the series:
$2\times 4+4\times 6+6\times 8+.....$

Answer 1

$2\times 4+4\times 6+6\times 8+.....$

In each term of sequence, first and second number forms an A.P. with common difference $2$ and $2$ respectively.

Let $T_n$ be the $n^{th}$ term of the given series then, $T_n=\{n_{th}\text{term of}2,4,6,...\}\times \{n^{th}\text{term of}4,6,8,...\}$

$=[2+(n-1)\times 2]\times [4+(n-1)\times 2]$

$=2n(2n+2)$

$\Rightarrow T_n=4n^2+4n\cdots \left(1\right)$

Put $n=20$

$\Rightarrow T_{20}=4\times (20{)}^2+4\times 20$

$=4\times 400+80=1680$

Let $S_n$ denote the sum of $n$ terms of the series whose $n^{th}$ terms is $T_n$, Then,

$S_n=\sum T_n$

$=\sum (4n^2+4n)$

$=4\sum n^2+4\sum n$

Upon simplification we get,

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

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

$=n(n+1)\left[\frac{4n+2+6}{3}\right]$

$=n(n+1)\left[\frac{(4n+8)}{3}\right]$

$\Rightarrow S_n=\frac{n}{3}(n+1)(4n+8)$

Put = 20

$\Rightarrow S_{20}=\frac{20}{3}(20+1)(4\times 20+8)$

$\Rightarrow S_{20}=12320$

Final answer:

$\therefore {20}^{th}$ term of series is 1680 and sum of 20 term series is 12320.