Question

The sum of the series $1·3^2+2·5^2+3·7^2+........$upto $20$ terms is

• A. $188090$
• B. $189080$
• C. $199080$
• D. None of these

Answer 1

The correct option is A

$188090$

Explanation for correct option

Let $S=1·3^2+2·5^2+3·7^2+........$

$$\begin{gathered} S=\sum _{r=1}^{n}r{(2r+1)}^2\Rightarrow S=\sum _{r=1}^n(4r^2+1+4r)\left(r\right) \\ \Rightarrow S=\sum _{r=1}^n(4r^3+r+4r^2) \\ \Rightarrow S=\sum _{r=1}^{n}4r^3+\sum _{r=1}^{n}4r^2+\sum _{r=1}^{n}r \\ \Rightarrow S=\underset{r=1}{\overset{n}{4\sum }}{r}^3+4\sum _{r=1}^n{r}^2+\sum _{r=1}^{n}r \\ \Rightarrow S=4{\left(\frac{n(n+1)}{2}\right)}^2+4\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2} \\ \Rightarrow S=4\left(\frac{n^2{(n+1)}^2}{4}\right)+\frac{2}{3}(2n^3+2n^2+n^2+n)+\frac{n^2+n}{2} \end{gathered}$$

as we know that $\sum _{r=1}^n{r}^3={\left(\frac{n(n+1)}{2}\right)}^2,\sum _{r=1}^n{r}^2=\frac{n(n+1)(2n+1)}{6},\sum _{r=1}^{n}r=\frac{n(n+1)}{2}$

Put $n=20$

$$\begin{gathered} S={\left(20\right)}^2{\left(21\right)}^2+\frac{2}{3}(2\times {\left(20\right)}^3+2{\left(20\right)}^2+{\left(20\right)}^2+20)+\frac{{\left(20\right)}^2+20}{2} \\ S=400\times 441+\frac{2}{3}(16000+800+400+20)+\frac{400+20}{2} \\ S=176400+11480+210 \\ S=188090 \end{gathered}$$

Hence, the correct option is $option\left(A\right)$.