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

The coefficient of $x^{32}$ in the expansion of $(x^4 - \frac{1}{x^3})^{15}$ is

The coefficient of $x^5$ in the expansion of $\left(2 x^3-\frac{1}{3 x^2}\right)^5$ is

Write the general term in the expansion of $(x^{2}-y)^{6}$.

In the binomial expansion of $(1+x)^{15}$,the coefficients of $x^{r}$ and $x^{r+3}$ are equal. Then,$r$ is

In the expansion of ${\left( \frac{1 + x}{1 - x} \right)^2}$,the coefficient of ${x^n}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module