Sum of Common Series

Edited By Komal Miglani | Updated on Oct 11, 2024 10:22 AM IST

If we add or subtract all the terms of a sequence we will get an expression, which is called a series. It is denoted by S_n. The sum of common series involves the sum of natural numbers, the Sum of the first n-odd natural number, the Sum of the first n-even natural number, the Sum of the squares of the first n-natural numbers, the Sum of the cube of the first n-natural numbers. In real life, we use the sum of series for calculating electrical circuits, population growth, and growth of bacteria.

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This Story also Contains

Sum of the first n-natural numbers
Sum of first n-odd natural number
Sum of first n-even natural number
Sum of the squares of first n-natural numbers
Derivation of Sum of the squares of first n-natural numbers
Sum of the cube of first n-natural numbers
Derivation of the Sum of the cube of first n-natural numbers
Solved Examples Based on the sum of common series
Summary

Sum of Common Series

In this article, we will cover the concept of the Sum of a common series. This category falls under the broader category of Sequence and series, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main Exam (from 2013 to 2023), a total of fifteen questions have been asked on this concept, including one in 2015, one in 2017, one in 2019, three in 2020, one in 2021, and one in 2023.

Sum of the first n-natural numbers

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

Sum of first n-odd natural number

The numbers which are not multiple of 2 are called odd numbers.

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

Sum of first n-even natural number

The numbers which are multiple of 2 are called even numbers.

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

Sum of the squares of first n-natural numbers

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

Derivation of Sum of the squares of first n-natural numbers

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

$\\\mathrm{Hence,\;by\;addition,}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\mathit{n^3}=3(1^2+2^2+3^2+........+n^2)-3(1+2+3+........+n)+n}$

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

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

Sum of the cube of first n-natural numbers

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

Derivation of the Sum of the cube of first n-natural numbers

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

Hence, by addition

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

Solved Examples Based on the sum of common series

Example 1: If $\frac{1^3+2^3+3^3+\cdots \text { up to } n \text { terms }}{1 \cdot 3+2 \cdot 5+3 \cdot 7+\cdots \text { up to } n \text { terms }}=\frac{9}{5}$ , then the value of n is [JEE MAINS 2022]

Solution

$\begin{aligned} & \frac{\left(\frac{n(n+1)}{2}\right)^2}{\sum r(2 r+1)} \\ & \Rightarrow \frac{\frac{n^2(n+1)^2}{4}}{\frac{2 n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}} \end{aligned}$

$\begin{aligned} \Rightarrow & \frac{\frac{n(n+1)}{4}}{\frac{2 n+1}{3}+\frac{1}{2}} \Rightarrow \frac{\frac{n(n+1)}{4}}{\frac{(4 n+5)}{6}}=\frac{9}{5} \\ \Rightarrow & \frac{3(n+1) n}{2(4 n+5)}=\frac{9}{5} \end{aligned}$

$\begin{aligned} & \Rightarrow 5 n^2+5 n=24 n+30 \\ & \Rightarrow 5 n^2-19 n-30=0 \\ & 5 n^2-25 n+6 n-30=0 \\ & (5 n+6)(n-5)=0 \\ & n=5 \end{aligned}$

Hence, the answer is 5.

Example 2: Let $S_n (x) = \log _{a^ {1/2}}x + \log _{a^ {1/3}}x + \log _{a^ {1/6}}x + \log _ _{a^ {1/11}}x + \log _ _{a^ {1/18}}x +\ldots$ upto n - terms, where a>1. If $S_{24} (x) =1093$ and $S_{12}(2x)= 265$ , then the value of a is equal to ______. [JEE MAINS 2021]

Solution

$S_n (x) = \log _{a^ {1/2}}x + \log _{a^ {1/3}}x + \log _{a^ {1/6}}x + \log _ _{a^ {1/11}}x + \log _ _{a^ {1/18}}x +\ldots$

$S_n (x) = 2\log _{a}x + 3\log _{a}x + 6\log _{a}x + 11\log _{a}x + 18\log _ {a}x +\ldots$

$S _{ n }( x )=(2+3+6+11+18+27+\ldots \ldots+ n \text { -terms }) \log _{ a } x$

Let

$\begin{aligned} &S_{1}=2+3+6+11+18+27+\ldots\ldots+T_{n} \\ &S_{1}=\;\;\;\;\;\;2+3+6+\ldots \ldots \ldots\ldots \ldots \ldots \ldots . .+T_{n} \end{array}$

$\\T _{ n }=2+1+3+5+\ldots \ldots+ n \text { terms } \\ T _{ n }=2+( n -1)^{2}$

$\\S_{1}=\Sigma T_{n}=2 n+\frac{(n-1) n(2 n-1)}{6} \\\\ \Rightarrow S_{n}(x)=\left(2 n+\frac{n(n-1)(2 n-1)}{6}\right) \log _{a} x$

$\\S _{24}( x )=1093(\text { Given }) \\ \\\log _{ a } x \left(48+\frac{23.24 .47}{6}\right)=1093 \\ \\\log _{ a } x =\frac{1}{4} \quad \ldots(1)$

$\\S _{12}(2 x )=265 \\ \\S _{12}(2 x )=265 \\ \\\log _{ a }(2 x )\left(24+\frac{11.12 .23}{6}\right)=265 \\ \\\log _{ a } 2 x =\frac{1}{2} \quad \ldots(2)$

$\begin{aligned} &(2)-(1)\\ &\log _{ a } 2 x -\log _{ a } x =\frac{1}{4}\\ &\log _{ a } 2=\frac{1}{4} \Rightarrow a =16 \end{aligned}$

Hence, the answer is 16

Example 3: The sum $\sum_{n=1}^{7}\frac{n(n+1)(2n+1)}{4}$ is equal to [JEE MAINS 2020]

Solution: Now,

$\begin{aligned} &\frac{1}{4}\left[\sum_{n=1}^{7}\left(2 n^{3}+3 n^{2}+n\right)\right]\\ &\frac{1}{4}\left[2\left(\frac{7.8}{2}\right)^{2}+3\left(\frac{7.8 .15}{6}\right)+\frac{7.8}{2}\right]\\ &\frac{1}{4}[2 \times 49 \times 16+28 \times 15+28]\\ &\frac{1}{4}[1568+420+28]=504 \end{aligned}$

Hence, the answer is 504

Example 4: The sum of all natural numbers 'n' such that $100<n<200 \;$ and $H.C.F (91,n)>1\;$ is [JEE MAINS 2019]

Solution: We know that the sum of the first n natural numbers is given by

$1+2+3+4+------+n= \frac{n(n+1)}{2}$

Now,

Natural number between 100 and 200

$=101,102,....199$

$As \,\,91 = 7\times 13, so for \; H.C.F \; (91,n)>\; 1$ ,

The number should either divide by 7 or divide by 13

Required sum = (sum of no.divisible by 7)+(sum of no divisible by 13)-(sum of no divisible by 91)

$=\sum_{r=1}^{14}(98+7r)+\sum_{r=1}^{8}(91+13r)-(182)$

$=98\times14+7\times\frac{14\times15}{2}+91\times8+13\times\frac{8\times9}{2}-182$

$=3121$

Hence, the answer is 3121

Example 5: The sum $\mathrm{1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+..........+\frac{1^3+2^3+3^3+......+15^3}{1+2+3+.......+15}\:-\:\frac{1}{2}\left(1+2+3+.......+15\right)}$ is equal to : [JEE MAINS 2019]

Solution: The sum of the first n natural numbers

$1+2+3+4+------+n= \frac{n(n+1)}{2}$

The sum of squares of first n natural numbers

$1^{2}+2^{2}+3^{2}+4^{2}+------+n^{2}= \frac{n(n+1)\left ( 2n+1 \right )}{6}$

The sum of cubes of first n natural numbers

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

Now,

$\mathrm{1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+..........+\frac{1^3+2^3+3^3+......+15^3}{1+2+3+.......+15}\:-\:\frac{1}{2}\left(1+2+3+.......+15\right)}$

$=\sum_{n=1}^{15}\frac{(\frac{n(n+1)}{2})^{2}}{\frac{n(n+1)}{2}}-\frac{1}{2}(\frac{15(1+15)}{2})$

$=\sum_{n=1}^{15}(\frac{n^{2}}{2}+\frac{n}{2})-\frac{1}{2}(\frac{15\times 16}{2})$

$=680-60$

$=620$

Hence, the answer is 620

Summary

The sum of common series refers to the calculation of the total value resulting from adding up the terms of a specific mathematical series. The study of sums of common series is fundamental in mathematics and its applications. These series not only deepen our understanding of mathematical structures but also play a critical role in fields such as physics, and engineering.