The coefficient of t^{12} in the expansion of | vedclass

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(C) The given expression is $(1 + t^2)^6(1 + t^6)(1 + t^{12})$.Expanding the last two terms: $(1 + t^6)(1 + t^{12}) = 1 + t^6 + t^{12} + t^{18}$.So,th

Vedclass · Accepted Answer

$22$

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(C) The given expression is $(1 + t^2)^6(1 + t^6)(1 + t^{12})$.Expanding the last two terms: $(1 + t^6)(1 + t^{12}) = 1 + t^6 + t^{12} + t^{18}$.So,the expression becomes $(1 + t^2)^6(1 + t^6 + t^{12} + t^{18})$.We need the coefficient of $t^{12}$ in the product $(1 + t^2)^6(1 + t^6 + t^{12} + t^{18})$.This is equal to:(Coefficient of $t^{12}$ in $(1 + t^2)^6 	imes 1$) + (Coefficient of $t^6$ in $(1 + t^2)^6 	imes t^6$) + (Coefficient of $t^0$ in $(1 + t^2)^6 	imes t^{12}$).Using the binomial expansion $(1 + t^2)^6 = \sum_{k=0}^{6} {^6C_k} (t^2)^k = \sum_{k=0}^{6} {^6C_k} t^{2k}$.$1$. Coefficient of $t^{12}$ in $(1 + t^2)^6$ is ${^6C_6} = 1$.$2$. Coefficient of $t^6$ in $(1 + t^2)^6$ is ${^6C_3} = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$.$3$. Coefficient of $t^0$ in $(1 + t^2)^6$ is ${^6C_0} = 1$.Summing these: $1 + 20 + 1 = 22$.

The coefficient of $t^{12}$ in the expansion of $(1 + t^2)^6(1 + t^6)(1 + t^{12})$ is:

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