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

^{12} C_{6} + 2

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^{12} C_{5}

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^{12} C_{6}

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^{12} C_{7}

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Solution

The correct option is B $^{12} C_{6} + 2$
We have, ${(1 + t^{2})}^{12} (1 + t^{12}) (1 + t^{24})$

$= (1 +^{12} C_{1} t^{2} +^{12} C_{2} t^{4} + \dots +^{12} c_{4} t^{8} + \dots +^{12} C_{10} t^{20} + \dots) (1 + t^{12} + t^{24} + t^{36})$

Therefore, coefficient of $t^{24}$ in

${(1 + t^{2})}^{12} (1 + t^{12}) (1 + t^{24}) =^{12} C_{6} + 2$

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Similar questions

t^{24}

{(1 - t)}^{12} {(t + t^{2})}^{12} + (t + t^{12}) (1 - t^{24})

t^{32}

{(1 + t^{2})}^{12} (1 + t^{12}) (1 + t^{24})

t^{32}

{(1 + t^{2})}^{12} (1 + t^{12}) (1 + t^{24})

Coefficient of $t^{24} i n (1 + t^{2})^{12} (1 + t^{12}) (1 + t^{24})$ is:

t^{12}

{(\frac{1 - t^{9}}{1 - t^{3}})}^{6}

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