Question

The sum of first $n$ terms of the series $1^2+\left(1^2+2^2\right)+\left(1^2+2^2+3^2\right)+\dots \dots$ is

• A. $\frac{n(n+1)(n+2)}{12}$
• B. $\frac{n(n+1{)}^2(n+2)}{12}$
• C. $\frac{n(n+1)(n+2{)}^2}{12}$
• D. $\frac{n(n+1{)}^2(n+2{)}^2}{12}$

Answer 1

Answer: (B)

Let the sum of first $n$ terms of the given series is $S=1^2+\left(1^2+2^2\right)+\left(1^2+2^2+3^2\right)+\dots$ upto $n$ terms
Here, first term has 1 term, second term has 2 terms, third terms has 3 and so on. Therefore, $n^{\text{th }}$ term will have $n$ terms
i.e., $T_n=1^2+2^2+\dots +n^2$
$\Rightarrow T_n=\Sigma n^2$
$\Rightarrow T_n=\frac{n(n+1)(2n+1)}{6}$
$\Rightarrow T_n=\frac{n\left(2n^2+n+2n+1\right)}{6}$
$\Rightarrow T_n=\frac{n\left(2n^2+3n+1\right)}{6}$
$\Rightarrow T_n=\frac{2n^3+3n^2+n}{6}$
Now, $S=\Sigma T_n=\frac{1}{6}\Sigma \left(2n^3+3n^2+n\right)$
$=\frac{1}{6}\left[2\Sigma n^3+3\Sigma n^2+\Sigma n\right]$
$=\frac{1}{6}\left[2{\left\{\frac{n(n+1)}{2}\right\}}^2+\frac{3n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right]$
$\left\{\because \Sigma n=\frac{n(n+1)}{2},\Sigma n^2=\frac{n(n+1)(2n+1)}{6},\Sigma n^3=\left[\frac{n(n+1)}{2}\right]\right\}$
$=\frac{1}{6}\times \frac{n{n}^3={\left[n+\frac{n(n+1)}{2}\right]}^2}{2}\left[\frac{2n(n+1)}{2}+\frac{2n+1}{1}+\frac{1}{1}\right]$
$=\frac{n(n+1)}{12}\times \left[n^2+n+2n+1+1\right]$
$=\frac{n(n+1)}{12}\left[n^2+3n+2\right]$
$=\frac{n(n+1)\left(n^2+2n+n+2\right)}{12}$
$=\frac{n(n+1)(n+2)(n+1)}{12}$
$=\frac{n(n+1{)}^2(n+2)}{12}$