SUMMATION FORMULA

Summation by Sigma(Σ) Operator

The summation of each term of a sequence or a series can be represented in a compact form, called summation or sigma notation. This summation is represented by the Greek capital letter, Sigma (Σ).

For example,

$\\\mathrm{\sum_{n=1}^{n=10} n\;,\;it\;means \;the \;sum\;of\;n\;terms\;when\;n\;varies\;from\;1\;to\;10}\\\mathrm{\sum_{n=1}^{n=10} n=1+2+3+4+5+6+7+8+9+10}$

If we have the formula for the rth term i.e. $A_r$ of the series, we can put the sum of n terms of the series in the form of sigma notation as

$S_{n}= a_{1}+ a_{2}+--------+ a_{n}$ = $\sum^n_{r=1} A_r$

Here, $A_r$ is called the general term of the series.

Thus, the sum of n terms of A.P. whose rth term is $A_r$ = a+ (r-1)*d; where a is the first term and d is the common difference is given by

$S_n = \sum^n_{r=1} A_r =\sum ^n_{r=1} a+ (r-1)*d$

In fact, we can put the sum of any series in the sigma notation if the formula for its rth term is known.

Properties of Sigma Notation

$\\\mathrm{1.\;\;\sum_{r=1}^{n}T_r=T_1+T_2+T_3+.......+T_n,\;where,\;T_r\;is\;the\;general\;term\;of\;the\;series.}\\\\\mathrm{2.\;\;\sum_{r=1}^{n}\left ( T_r\pm T_r' \right )=\sum_{r=1}^{n}T_r\pm\sum_{r=1}^{n}T_r'\;\;(sigma\;\;operator\;is\;distributive\;\;over\;addition\;and\;subtraction)}\\\\\mathrm{3.\;\;\sum_{r=1}^{n}T_rT_r'\neq\left ( \sum_{r=1}^{n}T_r \right )\left ( \sum_{r=1}^{n}T_r' \right )\;\;(sigma\;\;operator\;is\;not\;distributive\;\;over\;multiplication)}\\\\\mathrm{4.\;\;\sum_{r=1}^{n}\frac{T_r}{T_r'}\;\neq\;\frac{\sum_{r=1}^{n}T_r}{\sum_{r=1}^{n}T_r'}\;\;(sigma\;\;operator\;is\;not\;distributive\;\;over\;division})\\\\\mathrm{5.\;\;\sum_{r=1}^{n}aT_r=a\sum_{r=1}^{n}T_r\;\;\;\;(a\;is\;constant)}\\\\\mathrm{6.\;\;\sum_{j=1}^{n}\sum_{i=1}^{n}T_iT_j=\left ( \sum_{i=1}^{n}T_i \right )\left ( \sum_{j=1}^{n}T_j \right )\;\;\;(here\;i\;and\;j\;are\;independent)}$

Solved Examples Based on Summation by Sigma Operator

Example 1: Let $<a_n>$ be a sequence such that $a_1+a_2+\ldots+a_a=\frac{n^2+3 n}{(n+1)(n+2)}$, If $28 \sum_{k=1}^{10} \frac{1}{a_k}=p_1 p_2 p_3 \ldots p_m$, where $p_1, p_2 \ldots \ldots \mathrm{P}_w$ are the first m prime numbers, then m is equal to [JEE MAINS 2023]

Solution

$\begin{aligned} & a_n=S_n-S_{n-1}=\frac{n^2+3 n}{(n+1)(1+2)}-\frac{(n-1)(n+2)}{n(n+1)} \\ & \Rightarrow a_n=\frac{4}{n(n+1)(1+2)} \\ & \Rightarrow 28 \sum_{k-1}^{10} \frac{1}{a_k}=28 \sum_{k=1}^{10} \frac{k(k+1)(k+2)}{4} \\ & =\frac{7}{4} \sum_{k=1}^{10}(k(k+1)(k+2)(k+3)-(k-1) k(k+1)(k+2) \\ & =\frac{7}{4} .10 .11 .12 .13=2.3 .5 .7 .11 .13 \\ & \text { So } m=6 \\ & \end{aligned}$

Hence, the answer is 6

Example 2: Let $\sum_{n=0}^{\infty} \frac{n^3((2 n) !)+(2 n-1)(n !)}{(n !)((2 n) !)}=a e+\frac{b}{e}+c$, where $\text { a, b, c } \in \mathbb{Z}$ and $\mathrm{e}=\sum_{n=0}^{\infty} \frac{1}{n !}$ Then $a^2-b+c$ is equal to : [JEE MAINS 2023]

Solution

$\begin{aligned} & \text { Let } \sum_{n=0}^{\infty} \frac{n^3((2 n) !)+(2 n-1) n !}{(n !)((2 n) !)} \\ & =\sum_{n=0}^{\infty} \frac{n^3(2 n) !}{n !(2 n) !}+\frac{(2 n-1) n !}{n !(2 n) !} \\ & =S_1+S_2 \end{aligned}$

$\text { Let } \begin{aligned} S_1 & =\sum_{n=0}^{\infty} \frac{n^3(2 n) !}{n !(2 n) !}=\sum_{n=0}^{\infty} \frac{n^3}{n !}=\sum_{n=1}^{\infty} \frac{n^2}{(n-1) !} \\ & =\sum_{n=1}^{\infty} \frac{n^2-1+1}{(n-1) !} \\ & =\sum_{n=2}^{\infty} \frac{(n+1)}{(n-2) !}+\sum_{n=1}^{\infty} \frac{1}{(n-1) !} \end{aligned}$

$\begin{aligned} & =\sum_{n=2}^{\infty} \frac{(n-2)+3}{(n-2) !}+\sum_{n=1}^{\infty} \frac{1}{(n-1) !} \\ & =\sum_{n=3}^{\infty} \frac{1}{(n-3) !}+3 \sum_{n=2}^{\infty} \frac{1}{(n-2) !}+\sum_{n=1}^{\infty} \frac{1}{(n-1) !} \\ & S_1=e+3 e+e=5 e \\ & \because S_2=\sum_{n=0}^{\infty} \frac{(2 n-1) n !}{n !(2 n) !} \\ & =\sum_{n=0}^{\infty} \frac{2 n-1}{(2 n) !} \end{aligned}$

$\begin{aligned} & =\sum_{n=1}^{\infty} \frac{1}{(2 n-1) !}-\sum_{n=0}^{\infty} \frac{1}{(2 n) !} \\ & =\left(\frac{1}{1 !}+\frac{1}{3 !}+\frac{1}{5 !}+\ldots\right)-\left(1+\frac{1}{2 !}+\frac{1}{4 !}+\ldots .\right) \\ & =-1+\frac{1}{1 !}-\frac{1}{2 !}+\frac{1}{3 !}-\frac{1}{4 !}+\frac{1}{5 !}-\ldots \\ & =-\left(1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}+\ldots .\right) \\ & =-\mathrm{e}^{-1} \end{aligned}$

$\mathrm{S}_1+\mathrm{S}_2=5 \mathrm{e}-\frac{1}{\mathrm{e}}=\mathrm{ae}+\frac{\mathrm{b}}{\mathrm{e}}+\mathrm{c}$

Compare both sides

$\begin{aligned} & \mathrm{a}=5, \mathrm{~b}=-1, \mathrm{c}=0 \\ & \mathrm{a}^2-\mathrm{b}+\mathrm{c}=25+1+0=26 \end{aligned}$

Hence, the answer is 26.

Example 3: Let f(x) be a function such that $f(x+y)=f(x) \cdot f(y)$ for all $x, y \in N$ if $f(1)=3$ and $\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{f}(\mathrm{k})=3279$ then the value of n is. [JEE MAINS 2023]

Solution

$\begin{aligned} &\begin{aligned} & f(x+y)=f(x) \cdot f(y), x, y \in N \\ & f(2)=3^2 \end{aligned}\\ &\begin{aligned} f(3)=3^3 \quad & \therefore 3 \frac{\left[3^n-1\right]}{2}=3279 \\ & 3^n-1=1093 \times 2 \\ & 3^n-1=2186 \\ & 3^n=2187 \\ & n=7 \end{aligned} \end{aligned}$

Hence, the answer is 7

Example 4: Let $\mathrm{s}_1, \mathrm{~s}_2, \mathrm{~s}_3, \ldots \ldots, \mathrm{s}_{10}$ respectively be the sum to 12 terms of 10 A.P. s whose first terms are $1,2,3, \ldots, 10$ and the common differences are $1,3,5, \ldots \ldots \ldots, 19$ respectively. Then $\sum_{\mathrm{i}=1}^{10} \mathrm{~s}_{\mathrm{i}}$ is equal to [JEE MAINS 2023]

Solution

$\begin{aligned} & \mathrm{S}_{\mathrm{k}}=6(2 \mathrm{k}+(11)(2 \mathrm{k}-1)) \\ & \mathrm{S}_{\mathrm{k}}=6(2 \mathrm{k}+22 \mathrm{k}-11) \\ & \mathrm{S}_{\mathrm{k}}=144 \mathrm{k}-66 \\ \end{aligned}$

$\begin{aligned} & \sum_1^{10} \mathrm{~S}_{\mathrm{k}}=144 \sum_{\mathrm{k}=1}^{10} \mathrm{k}-66 \times 10 \\ & =144 \times \frac{10 \times 11}{2}-660 \\ & =7920-660 \\ & =7260 \end{aligned}$

Hence, the answer is 7260

Example 5: Let $[\alpha]$ denote the greatest integer $\leq \alpha$ . Then $[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\ldots .+[\sqrt{120}]$ is equal to _________. [JEE MAINS 2023]

Solution

$\begin{aligned} & \mathrm{S}=[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\ldots+[\sqrt{120}] \\ & {[\sqrt{1}] \rightarrow[\sqrt{3}]=1 \times 3} \\ & {[\sqrt{4}] \rightarrow[\sqrt{8}]=2 \times 5} \\ & {[\sqrt{9}] \rightarrow[\sqrt{15}]=3 \times 7} \\ & \vdots \\ & {[\sqrt{100}] \rightarrow[\sqrt{120}]=10 \times 21} \\ & \mathrm{~S}=1 \times 3+2 \times 5+3 \times 7+\ldots+10 \times 21 \\ & =\sum_{\mathrm{r}=1}^{10} \mathrm{r}(2 \mathrm{r}+1) \\ & =2 \sum_{\mathrm{r}=1}^{10} \mathrm{r}^2+\sum_{\mathrm{r}=1}^{10} \mathrm{r} \\ & =\frac{2 \times 10 \times 11 \times 21}{6}+\frac{10 \times 11}{2} \\ & =770+55 \\ & =825 \end{aligned}$

Hence, the answer is (825).