Find the sum of the series:

In the given question we need to find the sum of the series.

For that, first, we need to find the $\mathrm{n}^{\text {th }}$ term of the series so that we can use summation of the series with standard identities and get the required sum.

The series given is $(1 \times 2 \times 4)+(2 \times 3 \times 7)+(3 \times 4 \times 10)+\ldots$ to $n$ terms.

The series can be written as, $[(1 \times(1+1) \times(3 \times 1+1)),(2 \times(2+1) \times(3 \times 2+1)) \ldots(n \times$ $(n+1) \times(3 \times n+1)]$.

So, $\mathrm{n}^{\text {th }}$ term of the series,

$a_{n}=n(n+1)(3 n+1)$

$a_{n}=3 n^{3}+4 n^{2}+n$

Now, we need to find the sum of this series, Sn.

$\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{n}=1}^{\mathrm{n}} \mathrm{a}_{\mathrm{n}}$

$\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{n}=1}^{\mathrm{n}}\left(3 \mathrm{n}^{3}+4 \mathrm{n}^{2}+\mathrm{n}\right)$

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

$\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}$

II. Sum of squares of first $n$ natural numbers, $1^{2}+2^{2}+3^{2}+\ldots n^{2}$,

$\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}^{2}=\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}$

III. Sum of cubes of first $n$ natural numbers, $1^{3}+2^{3}+3^{3}+\ldots . n^{3}$,

$\sum_{k=1}^{n} k^{3}=\left(\frac{n(n+1)}{2}\right)^{2}$

IV. Sum of a constant k, N times,

$\sum_{k=1}^{N} k=N k$

So, for the given series, we need to find

$\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{n}=1}^{\mathrm{n}}\left(\mathrm{n}^{3}\right)+2 \sum_{\mathrm{n}=1}^{\mathrm{n}}\left(\mathrm{n}^{2}\right)+\sum_{\mathrm{n}=1}^{\mathrm{n}}(\mathrm{n})$

From, the above identities,

$S_{n}=3 \sum_{n=1}^{n}\left(n^{3}\right)+4 \sum_{n=1}^{n}\left(n^{2}\right)+\sum_{n=1}^{n}(n)$

$\mathrm{S}_{\mathrm{n}}=3\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)^{2}+4\left(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\right)+\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)$

$=\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)\left[\frac{3 \mathrm{n}(\mathrm{n}+1)}{2}+\frac{4(2 \mathrm{n}+1)}{3}+1\right]$

$\mathrm{S}_{\mathrm{n}}=\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)\left[\frac{9 \mathrm{n}^{2}+25 \mathrm{n}+14}{6}\right]$

So, Sum of the series $\mathrm{S}_{\mathrm{n}}=\left(\frac{\mathrm{n}(\mathrm{n}+1)}{12}\right)\left(9 \mathrm{n}^{2}+25 \mathrm{n}+14\right)$