(B) To find the sum of the coefficients $\sum_{r=0}^m a_r$,we substitute $x = 1$ into the given polynomial identity.Substituting $x = 1$ in the expres

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(B) To find the sum of the coefficients $\sum_{r=0}^m a_r$,we substitute $x = 1$ into the given polynomial identity.Substituting $x = 1$ in the expression $(1 + x) (1 + x + x^2) (1 + x + x^2 + x^3) \dots (1 + x + x^2 + \dots + x^n)$:$= (1 + 1) (1 + 1 + 1^2) (1 + 1 + 1^2 + 1^3) \dots (1 + 1 + 1^2 + \dots + 1^n)$$= (2) (3) (4) \dots (n + 1)$$= 1 \times 2 \times 3 \times 4 \dots \times (n + 1)$$= (n + 1)!$Thus,the sum of the coefficients is $(n + 1)!$.

Class 11 Mathematics Binomial Theorem Mix Examples-Binomial Theorem

If $(1 + x) (1 + x + x^2) (1 + x + x^2 + x^3) \dots (1 + x + x^2 + \dots + x^n) \equiv a_0 + a_1x + a_2x^2 + a_3x^3 + \dots + a_mx^m$,then $\sum_{r=0}^m a_r$ has the value equal to:

A
$n!$
B
$(n + 1)!$
C
$(n - 1)!$
D
None of these

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Mix Examples-Binomial Theorem

The coefficient of $x^{15}$ in the product $(1-x)(1-2x)(1-2^2x)(1-2^3x) \ldots (1-2^{15}x)$ is

MediumAP EAMCET 2018

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If $n$ is an odd positive integer and $(1+x+x^{2}+x^{3})^{n}=\sum_{r=0}^{3n} a_{r} x^{r}$,then $a_{0}-a_{1}+a_{2}-a_{3}+\ldots-a_{3n}$ is equal to

DifficultKCET 2011

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If $T_r = ^{2016}C_r x^{2016-r}$ for $r = 0, 1, 2, \dots, 2016$,then $(T_0 - T_2 + T_4 - \dots + T_{2016})^2 + (T_1 - T_3 + T_5 - \dots - T_{2015})^2$ is equal to-

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The sum of the series $2 \times 1 \times {}^{20}C_4 - 3 \times 2 \times {}^{20}C_5 + 4 \times 3 \times {}^{20}C_6 - 5 \times 4 \times {}^{20}C_7 + \dots + 18 \times 17 \times {}^{20}C_{20}$ is equal to:

DifficultJEE MAIN 2025

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Let $R=(5 \sqrt{5}+11)^{2 n+1}$ and $f=R-[R]$,where $[x]$ denotes the greatest integer less than or equal to $x$,then $R f=$

MediumAP EAMCET 2020

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If $(1 + x) (1 + x + x^2) (1 + x + x^2 + x^3) \dots (1 + x + x^2 + \dots + x^n) \equiv a_0 + a_1x + a_2x^2 + a_3x^3 + \dots + a_mx^m$,then $\sum_{r=0}^m a_r$ has the value equal to:

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The coefficient of $x^{15}$ in the product $(1-x)(1-2x)(1-2^2x)(1-2^3x) \ldots (1-2^{15}x)$ is

If $n$ is an odd positive integer and $(1+x+x^{2}+x^{3})^{n}=\sum_{r=0}^{3n} a_{r} x^{r}$,then $a_{0}-a_{1}+a_{2}-a_{3}+\ldots-a_{3n}$ is equal to

If $T_r = ^{2016}C_r x^{2016-r}$ for $r = 0, 1, 2, \dots, 2016$,then $(T_0 - T_2 + T_4 - \dots + T_{2016})^2 + (T_1 - T_3 + T_5 - \dots - T_{2015})^2$ is equal to-

The sum of the series $2 \times 1 \times {}^{20}C_4 - 3 \times 2 \times {}^{20}C_5 + 4 \times 3 \times {}^{20}C_6 - 5 \times 4 \times {}^{20}C_7 + \dots + 18 \times 17 \times {}^{20}C_{20}$ is equal to:

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