The coefficient of $t^{24}$ in the expansion of $\left( 1 + t^{2} \right)^{12}\left( 1 + t^{12} \right)(1 + t^{24})$ is

Question

Dr. Abubacker · Accepted Answer

(a)
We have, $\left( 1 + t^{2} ight)^{12}\left( 1 + t^{12} ight)(1 + t^{24})$
$= \left( 1 + \ ^{12}C_{1}t^{2} + \ ^{12}C_{2}t^{4} + ... + \ ^{12}C_{6}t^{12} + ... + \ ^{12}C_{12}t^{24} + ... ight)(1 + t^{12} + t^{24} + t^{36})$
$	herefore$ ${Coefficient\ of\ t^{24}\ in\ \left( 1 + t^{2} ight)}^{12}\left( 1 + t^{12} ight)(1 + t^{24})$
$= \ ^{12}C_{6} + \ ^{12}C_{12} + 1 = \ ^{12}C_{6} + 2$

Q. The coefficient of $t^{24}$ in the expansion of $\left( 1 + t^{2} \right)^{12}\left( 1 + t^{12} \right)(1 + t^{24})$ is

Explanation

Reviewed by Dr. Abubacker

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