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# Sum of Common Series - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Sum of Common Series (Part-2), Sum of Common Series (Part 3) are considered the most difficult concepts.

• Sum of Common Series - (Part 1) are considered the most asked concepts.

• 11 Questions around this concept.

## Solve by difficulty

The number of common terms in the progressions $4,9,14,19, \ldots \ldots$, up to $25^{\text {th }}$ term and $3,6,9,12, \ldots \ldots$, up to $37^{\text {th }}$ term is :

## Concepts Covered - 3

Sum of Common Series - (Part 1)

Sum of Common Series

1. Sum of the first n-natural numbers

$\\\mathrm{1+2+3+4+5+........+n=\frac{n(n+1)}{2}}\\\mathrm{Use\;the \;formula\;,\;sum\;of\;n-term\;of\;an\;AP}\\\mathrm{S_n=\frac{n}{2}(a+l);\;\;where,\;a=1,\;\;l=n\;\;and\;number \;of\;term\;is\;n}\\\mathrm{S_n=\frac{n}{2}(1+n)}\\\mathrm{\Rightarrow \sum _{n=1}^nn=\frac{n(n+1)}{2}}$

1. Sum of first n-odd natural number

$\\\mathrm{1+3+5+7+.......upto\;n\;term=\frac{n}{2}\left ( 2\cdot1+(n-1)\cdot2 \right )}\\\mathrm{\Rightarrow \sum (2n-1)=n^2}$

Sum of Common Series (Part-2)

Sum of Common Series

1. Sum of first n-even natural number

2 + 4 + 6 + 8 + ……… = n/2 [2 x 2 + (n-1)2 ] = n(n+1)

1. Sum of the squares of first n-natural numbers

$\\\mathrm{1^2+2^2+3^2+4^2+...........+n^2=\frac{n(n+1)(2n+1)}{6}}$

Proof:

$\\\mathrm{we\;have,\;\mathit{n^3-(n-1)^3=3n^2-3n+1};\;by\;changing\;\mathit{n}\;to\;\mathit{(n-1)}\;,we\;get}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\mathit{(n-1)^3-(n-2)^3=3(n-1)^2-3(n-1)+1}}\\\mathit{\;\;\;\;\;\;\;\;\;\;\;\;(n-2)^3-(n-3)^3=3(n-2)^2-3(n-2)+1}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\cdot}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\cdot}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;3^3-2^3=3\times3^2-3\times 3+1}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;2^3-1^3=3\times2^2-3\times 2+1}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;1^3-0^3=3\times1^2-3\times 1+1}$$\\\mathrm{Hence,\;by\;addition,}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\mathit{n^3}=3(1^2+2^2+3^2+........+n^2)-3(1+2+3+........+n)+n}$

$\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=3S-\frac{3n(n+1)}{2}+n}$

$\\\mathrm{3S=\mathit{n^3-n+\frac{3n(n+1)}{2}}}\\\mathrm{\;\;\;\;\;=\mathit{n(n+1)(n-1+\frac{3}{2})}}\\\mathrm{\Rightarrow S=\mathit{\frac{n(n+1)(2n+1)}{6}}}$

Sum of Common Series (Part 3)

Sum of Common Series

1. Sum of the cube of first n-natural numbers

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

Proof:

$\\\mathrm{We\;have,}\\\;\;\;\;\;n^4-(n-1)^4=4n^3-6n^2+4n-1\\\begin{array}{l}{(n-1)^{4}-(n-2)^{4}=4(n-1)^{3}-6(n-1)^{2}+4(n-1)-1} \\ {(n-2)^{4}-(n-3)^{4}=4(n-2)^{3}-6(n-2)^{2}+4(n-2)-1} \\ {\vdots} \\ {3^{4}-2^{4}=4 \times 3^{3}-6 \times 3^{2}+4 \times 3-1} \\ {2^{4}-1^{4}=4 \times 2^{3}-6 \times 2^{2}+4 \times 2-1} \\ {1^{4}-0^{4}=4 \times 1^{3}-6 \times 1^{2}+4 \times 1-1}\end{array}$

\begin{aligned} n^{4} &=4 S-6\left(1^{2}+2^{2}+\cdots+n^{2}\right)+4(1+2+\cdots+n)-n \\ 4 S &=n^{4}+n+6\left(1^{2}+2^{2}+\cdots+n^{2}\right)-4(1+2+\cdots+n) \\ &=n^{4}+n+n(n+1)(2 n+1)-2 n(n+1) \\ &=n(n+1)\left(n^{2}-n+1+2 n+1-2\right) \\ &=n(n+1)\left(n^{2}+n\right) \\ S &=\frac{n^{2}(n+1)^{2}}{4}=\left\{\frac{n(n+1)}{2}\right\}^{2} \end{aligned}

## Study it with Videos

Sum of Common Series - (Part 1)
Sum of Common Series (Part-2)
Sum of Common Series (Part 3)

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## Books

### Reference Books

#### Sum of Common Series (Part-2)

Mathematics for Joint Entrance Examination JEE (Advanced) : Algebra

Page No. : 5.24

Line : 41

#### Sum of Common Series (Part 3)

Mathematics for Joint Entrance Examination JEE (Advanced) : Algebra

Page No. : 5.25

Line : 1