Question

The sum of$n$ terms of the infinite series $1.3^2+2.5^2+3.7^2+\dots$ is

• A. $\frac{n}{6}(n+1)(6n^2+14n+7)$
• B. $\frac{n}{6}(n+1)(2n+1)(3n+1)$
• C. $4n^3+4n^2+n$
• D. none of these

Answer 1

The correct option is D

none of these

Explanation for the correct option:

Find the required sum.

Given: series $1.3^2+2.5^2+3.7^2+\dots n$ terms

Here,$1,2,3,\dots ..$are in AP

$a=1,d=1$

$$\begin{gathered} t_n=a+(n-1)d \\ =1+(n-1)1 \\ =n \end{gathered}$$

And $3,5,7,\dots ..$ are in AP

$a=3,d=2$

$$\begin{gathered} t_n=a+(n-1)d \\ =3+(n-1)2 \\ =2n+1 \end{gathered}$$

$\therefore n^{th}$term of given series $=n{(2n+1)}^2$

$=4n^3+4n^2+n$

Now, sum of $n$term $S_n=\sum (4n^3+4n^2+n)$

$$\begin{gathered} =4\sum n^3+4\sum n^2+\sum n \\ =4\left[\frac{n^2{\left(n+1\right)}^2}{4}\right]+4\left[\frac{n\left(n+1\right)(2n+1)}{6}\right]+\left[\frac{n(n+1)}{2}\right] \\ =n^2{\left(n+1\right)}^2+\frac{2n\left(n+1\right)(2n+1)}{3}+\frac{n(n+1)}{2} \end{gathered}$$

Hence, option (D) is the correct answer.