Binomial Theorem 1 Question 19
21. Coefficient of $t^{24}$ in $\left(1+t^{2}\right)^{12}\left(1+t^{12}\right)\left(1+t^{24}\right)$ is
(a) ${ }^{12} C _6+3$
(b) ${ }^{12} C _6+1$
(c) ${ }^{12} C _6$
(d) ${ }^{12} C _6+2$
(2003, 1M)
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Solution:
- Here, Coefficient of $t^{24}$ in ${\left(1+t^{2}\right)^{12}\left(1+t^{12}\right)\left(1+t^{24}\right) }$ $=$ Coefficient of $t^{24}$ in ${\left(1+t^{2}\right)^{12} \cdot\left(1+t^{12}+t^{24}+t^{36}\right) }$ $=$ Coefficient of $t^{24}$ in
$$ {\left(1+t^{2}\right)^{12}+t^{12}\left(1+t^{2}\right)^{12}+t^{24}\left(1+t^{2}\right)^{12} } $$
[neglecting $t^{36}\left(1+t^{2}\right)^{12}$ ]
$=$ Coefficient of $t^{24}=\left({ }^{12} C _{12}+{ }^{12} C _6+{ }^{12} C _0\right)=2+{ }^{12} C _6$
