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

Here the nth term of the series is $T_n=2n(2n+2)$ Thus the 20th term will be $T_{20}=2\times 20(2\times 20+2)=1680$ The infinite series can be written as : $2\times 4+4\times 6+6\times 8+...={\sum }_{n-1}^{20}2n(2n+1)$ Therefore the sum up to 20${}^{th}$ term will be =${\sum }_{n-1}^{20}2n(2n+1)={\sum }_{n-1}^{20}4n^2+{\sum }_{n-1}^{20}4n=4{\sum }_{n-1}^{20}{n}^2+4{\sum }_{n-1}^{20}n=4.\frac{20(20+1)(2.20+1)}{6}+4.\frac{20(20+1)}{2}$ =12320