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

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(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

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$^{12}C_6+2$

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(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

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