Thank you for reporting, we will resolve it shortly
Q.
The sum of first $n$ terms of the series $1^2+\left(1^2+2^2\right)+\left(1^2+2^2+3^2\right)+\ldots \ldots$ is
Sequences and Series
Solution:
Let the sum of first $n$ terms of the given series is $S=1^2+\left(1^2+2^2\right)+\left(1^2+2^2+3^2\right)+\ldots$ 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., $T_n=1^2+2^2+\ldots+n^2$
$\Rightarrow T_n=\Sigma n^2$
$\Rightarrow T_n=\frac{n(n+1)(2 n+1)}{6} $
$ \Rightarrow T_n=\frac{n\left(2 n^2+n+2 n+1\right)}{6}$
$ \Rightarrow T_n=\frac{n\left(2 n^2+3 n+1\right)}{6}$
$ \Rightarrow T_n=\frac{2 n^3+3 n^2+n}{6}$
Now, $S =\Sigma T_n=\frac{1}{6} \Sigma\left(2 n^3+3 n^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{3 n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}\right] $
$ \left\{\because \Sigma n=\frac{n(n+1)}{2}, \Sigma n^2=\frac{n(n+1)(2 n+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{2 n(n+1)}{2}+\frac{2 n+1}{1}+\frac{1}{1}\right]$
$ =\frac{n(n+1)}{12} \times\left[n^2+n+2 n+1+1\right]$
$=\frac{n(n+1)}{12}\left[n^2+3 n+2\right]$
$=\frac{n(n+1)\left(n^2+2 n+n+2\right)}{12}$
$=\frac{n(n+1)(n+2)(n+1)}{12}$
$=\frac{n(n+1)^2(n+2)}{12}$