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 $5+7+13+31+85+\ldots .+\mathrm{n}$ 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 an, 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.

$\mathrm{S}_{\mathrm{n}}=5+7+13+31+85+\ldots . .+\mathrm{an} \rightarrow(1)$

$\mathrm{S}_{\mathrm{n}}=5+7+13+31+85+\ldots .+\mathrm{an} \rightarrow(2)$

$(1)-(2) \rightarrow 0=5+2+6+18+54+\ldots . .+\left(a_{n}-a_{n-1}\right)-a_{n}$

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

$a_{n}=5+2+6+18+54+\ldots$

In the resulting series obtained, starting from $2,6,18 \ldots$ forms a GP.

So, the $\mathrm{n}^{\text {th }}$ term forms a GP, with the first term, $\mathrm{a}=2$; common ratio, $\mathrm{r}=3$.

The required $\mathrm{n}^{\text {th }}$ term of the series is the same as the sum of $\mathrm{n}$ terms of GP and $5 .$

The GP is $2+6+18+54+\ldots(\mathrm{n}-1)$ terms.

Sum of n terms of a GP, $\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{a}\left(\mathrm{r}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}$

Sum of ( $n-1$ ) terms of a GP, $S_{n}=\frac{a\left(r^{n-1}-1\right)}{r-1}$

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

$=3^{n-1}-1$

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

So, $\mathrm{n}^{\text {th }}$ term of the series, $a_{n}=\frac{3^{n}}{3}-1+5=\frac{3^{n}}{3}+4$

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

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

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

Note:

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

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

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,

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

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

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

From, the above identities,

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

The first term in (a) is a GP, with the first term, a = 3 and common ratio, r = 3.

Sum of n terms of GP, $S_{n}^{1}=\frac{a\left(r^{n}-1\right)}{r-1}$

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

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

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

$\mathrm{S}_{\mathrm{n}}=\left[\frac{8 \mathrm{n}+3^{\mathrm{n}}-1}{2}\right]$

So, Sum of the series, $\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[8 \mathrm{n}+3^{\mathrm{n}}-1\right]$