# Find the sum of the series:

Question:

Find the sum of the series:

$(1 \times 2 \times 3)+(2 \times 3 \times 4)+(3 \times 4 \times 5)+\ldots$ to $n$ terms

Solution:

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 $(1 \times 2 \times 3)+(2 \times 3 \times 4)+(3 \times 4 \times 5)+\ldots$ to $n$ terms.

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

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

$a_{n}=n(n+1)(n+2)$

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

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

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

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

Note:

I. Sum of first $n$ natural numbers, $1+2+3+\ldots n$,

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

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,

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

From, the above identities,

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

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

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

$=\left(\frac{n(n+1)}{2}\right)\left[\frac{n^{2}+5 n+6}{2}\right]$

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

So, Sum of the series, $S_{n}=\left(\frac{n(n+1)(n+2)(n+3)}{4}\right)$