Find the ${{n}^{th}}$ term of the series $1+3+7+13+21+......$ and hence find the sum of first n terms?
Last updated date: 27th Mar 2023
•
Total views: 207.3k
•
Views today: 1.84k
Answer
207.3k+ views
Hint: 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
$\begin{align}
& S=1+3+7+13+21+......+{{t}_{n}} \\
& \underline{S=1+3+7+13+21+......+{{t}_{n}}} \\
& 0=1+\left( 3-1 \right)+\left( 7-3 \right)+\left( 13-7 \right)+......+\left( {{t}_{n}}-{{t}_{n-1}} \right)-{{t}_{n}} \\
\end{align}$
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
\[\begin{align}
& \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}-\dfrac{n\left( n+1 \right)}{2}+n \\
& =\dfrac{n\left( n+1 \right)\left( 2n+1-3 \right)}{6}+n \\
& =\dfrac{n\left( {{n}^{2}}-1 \right)}{3}+n \\
& =\dfrac{n}{3}\left( {{n}^{2}}+2 \right) \\
\end{align}\]
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.
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
$\begin{align}
& S=1+3+7+13+21+......+{{t}_{n}} \\
& \underline{S=1+3+7+13+21+......+{{t}_{n}}} \\
& 0=1+\left( 3-1 \right)+\left( 7-3 \right)+\left( 13-7 \right)+......+\left( {{t}_{n}}-{{t}_{n-1}} \right)-{{t}_{n}} \\
\end{align}$
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
\[\begin{align}
& \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}-\dfrac{n\left( n+1 \right)}{2}+n \\
& =\dfrac{n\left( n+1 \right)\left( 2n+1-3 \right)}{6}+n \\
& =\dfrac{n\left( {{n}^{2}}-1 \right)}{3}+n \\
& =\dfrac{n}{3}\left( {{n}^{2}}+2 \right) \\
\end{align}\]
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.
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE
