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    SUMMATION FORMULA - Practice Questions & MCQ

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

    Quick Facts

    • Summation by Sigma Operator is considered one the most difficult concept.

    • 19 Questions around this concept.

    Solve by difficulty

    If $f(x)=\frac{2^x}{2^x+\sqrt{2}}, x \in R$, then $\sum_{k=1}^{81} f\left(\frac{k}{82}\right)$ is equal to:

    Concepts Covered - 1

    Summation by Sigma Operator

    Summation by Sigma( $\Sigma$ ) 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 ( $\Sigma$ ).

    For example,
    $\mathrm{n}=10$
    $\sum_{n=1}^n n$, it means the sum of $n$ terms when $n$ varies from 1 to 10

    $
    \sum_{n=1}^{n=10} n=1+2+3+4+5+6+7+8+9+10
    $
    Properties of Sigma Notation
    1. $\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{T}_{\mathrm{r}}=\mathrm{T}_1+\mathrm{T}_2+\mathrm{T}_3+\ldots \ldots+\mathrm{T}_{\mathrm{n}}$, where, $\mathrm{T}_{\mathrm{r}}$ is the general term of the series.
    2. $\sum_{\mathrm{r}=1}^{\mathrm{n}}\left(\mathrm{T}_{\mathrm{r}} \pm \mathrm{T}_{\mathrm{r}}^{\prime}\right)=\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{T}_{\mathrm{r}} \pm \sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{T}_{\mathrm{r}}^{\prime}$ (sigma operator is distributive over addition and subtraction)
    3. $\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{T}_{\mathrm{r}} \mathrm{T}_{\mathrm{r}}^{\prime} \neq\left(\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{T}_{\mathrm{r}}\right)\left(\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{T}_{\mathrm{r}}^{\prime}\right)$ (sigma operator is not distributive over multiplication)
    4. $\sum_{\mathrm{r}=1}^{\mathrm{n}} \frac{\mathrm{T}_{\mathrm{r}}}{\mathrm{T}_{\mathrm{r}}^{\prime}} \neq \frac{\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{T}_{\mathrm{r}}}{\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{T}_{\mathrm{r}}^{\prime}}$ (sigma operator is not distributive over division)
    5. $\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{a} \mathrm{T}_{\mathrm{r}}=\mathrm{a} \sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{T}_{\mathrm{r}} \quad$ ( a is constant)
    6. $\sum_{j=1}^n \sum_{i=1}^n T_i T_j=\left(\sum_{i=1}^n T_i\right)\left(\sum_{j=1}^n T_j\right) \quad$ (here i and j are independent)

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    Summation by Sigma Operator

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