# Find the sum of the series:

Question:

Find the sum of the series:

1 + 5 + 12 + 22 + 35 +... 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 $1+5+12+22+35 \ldots$ to $n$ terms.

This question can be solved by the method of difference.

Note:

Consider a sequence $a_{1}, a_{2}, a_{3} \ldots$ such that the Sequence $a_{2}-a_{1}, a_{3}-a_{2} \ldots$ is either an. A.P. or a G.P.

The $\mathrm{n}^{\text {th }}$ term, of this sequence, is obtained as follows:

$S=a_{1}+a_{2}+a_{3}+\ldots+a_{n-1}+a_{n} \rightarrow(1)$

$S=a_{1}+a_{2}+\ldots+a_{n-2}+a_{n-1}+a_{n} \rightarrow(2)$

Subtracting (2) from (1),

We get, $a_{n}=a_{1}+\left[\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+\ldots\left(a_{n}-a_{n-1}\right)\right]$

Since the terms within the brackets are either in an A.P. or a G.P, we can find the value of $a_{n}$ the $n^{\text {th }}$ term.

Thus, we can find the sum of the n terms of the sequence as,

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

So

By using the method of difference, we can find the $\mathrm{n}^{\text {th }}$ term of the expression.

$\mathrm{S}_{\mathrm{n}}=1+5+12+22+35+\ldots . .+\mathrm{a}_{\mathrm{n}} \rightarrow(1)$

$\mathrm{S}_{\mathrm{n}}=1+5+12+22+35+\ldots .+\mathrm{an} \rightarrow(2)$

$(1)-(2) \rightarrow 0=1+4+7+10+\ldots \ldots-a_{n}$

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

$a_{n}=1+4+7+10+\ldots$

So, the $\mathrm{n}^{\text {th }}$ term form an AP, with the first term, $\mathrm{a}=1$; common difference, $\mathrm{d}=3$.

The required $\mathrm{n}^{\text {th }}$ term of the series is the same as the sum of $\mathrm{n}$ terms of AP.

Sum of $n$ terms of an $A P, S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \times 1+3(\mathrm{n}-1)]$

$=\frac{\mathrm{n}}{2}[3 \mathrm{n}-1]=\frac{3 \mathrm{n}^{2}-\mathrm{n}}{2}$

So, $\mathrm{n}^{\text {th }}$ term of the series, $\mathrm{a}_{\mathrm{n}}=\frac{3 \mathrm{n}^{2}-\mathrm{n}}{2}$

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

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

$S_{n}=\sum_{n=1}^{n} \frac{3 n^{2}-n}{2}$

Note:

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

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

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

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

IV. Sum of a constant k, N times,

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

So, for the given series, we need to find,

$\mathrm{S}_{\mathrm{n}}=\frac{1}{2}\left[3 \sum_{\mathrm{n}=1}^{\mathrm{n}}\left(\mathrm{n}^{2}\right)-\sum_{\mathrm{n}=1}^{\mathrm{n}}(\mathrm{n})\right]$

From, the above identities,

$\mathrm{S}_{\mathrm{n}}=\frac{1}{2}\left[3 \sum_{\mathrm{n}=1}^{\mathrm{n}}\left(\mathrm{n}^{2}\right)-\sum_{\mathrm{n}=1}^{\mathrm{n}}(\mathrm{n})\right]$

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

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

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

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