Expansion?

Probability Level 2

Find the coefficient of $t^{24}$ in the expansion of $(1+t^{2})^{12}(1+t^{12})(1+t^{24})$ .

Notation : $\dbinom MN$ denotes the binomial coefficient , $\dbinom MN = \dfrac{M!}{N!(M-N)!}$ .

\dbinom{12}{6}

\dbinom{12}{5}

\dbinom{12}{6}+2

\dbinom{12}{7}

2 solutions

Gaurav Chahar
May 10, 2016

Just focus on binomial expansion of first bracket and then collect t*24 from the expression

Thanks for sharing your approach, Gaurav. It would be great if you could include more details about you solved the problem - it would help the reader understand the solution better. :)

Pranshu Gaba - 5 years, 1 month ago

Sabhrant Sachan
May 10, 2016

We can say that $(1+t^2)^{12}=\displaystyle\sum_{n=0}^{12}\dbinom{12}{n}t^{2n}$

Now $(1+t^{2})^{12}(1+t^{12})(1+t^{24}) \\ \implies (\displaystyle\sum_{n=0}^{12}\dbinom{12}{n}t^{2n})(1+t^{12}+t^{24}+t^{36}) \\ \displaystyle\sum_{n=0}^{12}\dbinom{12}{n}t^{2n}+\displaystyle\sum_{n=0}^{12}\dbinom{12}{n}t^{2n+12}+\displaystyle\sum_{n=0}^{12}\dbinom{12}{n}t^{2n+24}+\displaystyle\sum_{n=0}^{12}\dbinom{12}{n}t^{2n+36}$

To get $t^{24}$ , Possible value of $n =12,6,0$

$\implies \dbinom{12}{12}+\dbinom{12}{6}+\dbinom{12}{0} \\ \color{#3D99F6}{\boxed{\text{Ans = }\dbinom{12}{6}+2}}$

Expansion?

2 solutions

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