The coefficient of x^{24} in the expansion of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given expression is $(1+x^2)^{12}(1+x^{12})(1+x^{24})$.First,simplify the product $(1+x^{12})(1+x^{24}) = 1 + x^{12} + x^{24} + x^{36}$.Now,th

Vedclass · Accepted Answer

$^{12}C_6+2$

Vedclass · Answer

(B) The given expression is $(1+x^2)^{12}(1+x^{12})(1+x^{24})$.First,simplify the product $(1+x^{12})(1+x^{24}) = 1 + x^{12} + x^{24} + x^{36}$.Now,the expression becomes $(1+x^2)^{12}(1 + x^{12} + x^{24} + x^{36})$.Using the binomial expansion,$(1+x^2)^{12} = \sum_{r=0}^{12} {}^{12}C_r (x^2)^r = \sum_{r=0}^{12} {}^{12}C_r x^{2r}$.We need the coefficient of $x^{24}$ in the product $(\sum_{r=0}^{12} {}^{12}C_r x^{2r})(1 + x^{12} + x^{24} + x^{36})$.This is obtained by:$1$. The term from the sum where $2r = 24$,which is ${}^{12}C_{12} x^{24} 	imes 1 = {}^{12}C_{12} x^{24}$.$2$. The term from the sum where $2r = 12$,which is ${}^{12}C_6 x^{12} 	imes x^{12} = {}^{12}C_6 x^{24}$.$3$. The term from the sum where $2r = 0$,which is ${}^{12}C_0 x^0 	imes x^{24} = 1 	imes x^{24} = 1 x^{24}$.Summing these coefficients: ${}^{12}C_{12} + {}^{12}C_6 + 1 = 1 + {}^{12}C_6 + 1 = {}^{12}C_6 + 2$.

The coefficient of $x^{24}$ in the expansion of $(1+x^2)^{12}(1+x^{12})(1+x^{24})$ is

Similar Questions

If for positive integers $r > 1, n > 2$,the coefficients of the $(3r)^{th}$ and $(r + 2)^{th}$ powers of $x$ in the expansion of $(1 + x)^{2n}$ are equal,then $n$ is equal to

Show that the middle term in the expansion of $(1+x)^{2n}$ is $\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} 2^n x^n$,where $n$ is a positive integer.

The term independent of $x$ in ${\left[ {\sqrt{\frac{x}{3}} + \frac{{\sqrt{3}}}{{{x^2}}}} \right]^{10}}$ is

In the expansion of $(x^2 - 2x)^{10}$,the coefficient of $x^{16}$ is

Find the middle terms in the expansions of $\left(3-\frac{x^{3}}{6}\right)^{7}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module