Question

Find the ${{n}^{th}}$ term of the series $1+3+7+13+21+......$ and hence find the sum of first n terms?

Answer 1

We first try to find the ${{n}^{th}}$ term of the series using the subtraction form of the terms. In that case, we subtract shifting one term on the right side. We find the term form in general and then using the formulas of $\sum\limits_{p=1}^{n}{{{p}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6},\sum\limits_{p=1}^{n}{p}=\dfrac{n\left( n+1 \right)}{2}$ we find sum.

Complete step by step solution:
We have been given $1+3+7+13+21+......$. This is AGM progression.
We assume its ${{n}^{th}}$ term as ${{t}_{n}}$. So, $1+3+7+13+21+......+{{t}_{n}}$
We assume the sum as $S=1+3+7+13+21+......+{{t}_{n}}$.
We apply one particular trick to subtract S from S in a particular way taking crosswise subtraction as
$\begin{aligned} & S=1+3+7+13+21+......+{{t}_{n}} \\\ & \underline{S=1+3+7+13+21+......+{{t}_{n}}} \\\ & 0=1+\left( 3-1 \right)+\left( 7-3 \right)+\left( 13-7 \right)+......+\left( {{t}_{n}}-{{t}_{n-1}} \right)-{{t}_{n}} \\\ \end{aligned}$
The series has $n$ terms. The simplified form is $1+2+4+6+......+\left( {{t}_{n}}-{{t}_{n-1}} \right)-{{t}_{n}}=0$.
So, ${{t}_{n}}=1+2+4+6+......+\left( {{t}_{n}}-{{t}_{n-1}} \right)$ and it has n terms. This series is an AP.
We use the known summation forms like $\sum\limits_{p=1}^{n}{{{p}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6},\sum\limits_{p=1}^{n}{p}=\dfrac{n\left( n+1 \right)}{2}$.
The sum will be ${{t}_{n}}=1+2+4+6+......+\left( {{t}_{n}}-{{t}_{n-1}} \right)=1+\sum\limits_{r=1}^{n-1}{2r}=1+n\left( n-1 \right)$.
The terms will be in the form of ${{t}_{n}}=1+n\left( n-1 \right)={{n}^{2}}-n+1$ putting values $n=1,2,3...$
So, we get $S=1+3+7+13+21+......+{{t}_{n}}=\sum{{{n}^{2}}-n+1}=\sum\limits_{k=1}^{n}{{{k}^{2}}-k+1}$
The summation gives us $S=\sum\limits_{k=1}^{n}{{{k}^{2}}-k+1}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}-\dfrac{n\left( n+1 \right)}{2}+n$.
Now we need to simplify the summation and get

& \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}-\dfrac{n\left( n+1 \right)}{2}+n \\\ & =\dfrac{n\left( n+1 \right)\left( 2n+1-3 \right)}{6}+n \\\ & =\dfrac{n\left( {{n}^{2}}-1 \right)}{3}+n \\\ & =\dfrac{n}{3}\left( {{n}^{2}}+2 \right) \\\ \end{aligned}$$ **Therefore, the ${{n}^{th}}$ term of the series $1+3+7+13+21+......$ is ${{t}_{n}}={{n}^{2}}-n+1$ and the sum of first n terms is $$\dfrac{n}{3}\left( {{n}^{2}}+2 \right)$$.** **Note:** This special form of subtraction is done to get the ${{t}_{n}}$ in negative form with the number of terms in that subtraction being equal to the number of terms in the main series. The summation forms of $$\sum\limits_{p=1}^{n}{{{p}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6},\sum\limits_{p=1}^{n}{p}=\dfrac{n\left( n+1 \right)}{2}$$ comes from that too.