# Find the sum of the series:

Question:

Find the sum of the series:

$\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+\ldots$ to n terms

Solution:

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

For that, firs, 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 $\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+\cdots$ to $\mathrm{n}$ terms

The series can be written as,

$\frac{1^{3}}{1}, \frac{1^{3}+2^{3}}{1+3}, \frac{1^{3}+2^{3}+3^{3}}{1+3+5}, \cdots \frac{1^{3}+2^{3}+3^{3}+\cdots+n^{3}}{1+3+5+\cdots+(2 n-1)}$

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

$a_{n}=\frac{1^{3}+2^{3}+3^{3}+\cdots n^{3}}{1+3+5+\cdots+(2 n-1)}$

The denominator of ' $a_{n}$ ' forms an $A P$ with first term $a=1$, last term $=2 n-1$ and common difference, $d=2$.

Now, Sum of the AP, $\mathrm{T}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$

$=\frac{\mathrm{n}}{2}[2+(\mathrm{n}-1) 2]=\mathrm{n}^{2}$

$a_{n}=\frac{\left(\frac{n(n+1)}{2}\right)^{2}}{n^{2}}=\frac{(n+1)^{2}}{4}$

$a_{n}=\frac{n^{2}+2 n+1}{4}$

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

$S_{n}=\sum_{n=1}^{n} a_{n}$

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

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, $1^{2}+2^{2}+3^{2}+\ldots n^{2}$,

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

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,

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

From, the above identities,

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

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

$=\left(\frac{n}{4}\right)\left[\frac{(n+1)(2 n+1)}{6}+(n+1)+1\right]$

$\mathrm{S}_{\mathrm{n}}=\left(\frac{\mathrm{n}}{4}\right)\left[\frac{2 \mathrm{n}^{2}+9 \mathrm{n}+13}{6}\right]$

So, Sum of the series, $S_{n}=\left(\frac{n}{24}\right)\left(2 n^{2}+9 n+13\right)$