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Maths-

General

Easy

Question

The coefficient of $t to the power of 24 end exponent$ in the expansion of $open parentheses 1 plus t to the power of 2 end exponent close parentheses to the power of 12 end exponent open parentheses 1 plus t to the power of 12 end exponent close parentheses left parenthesis 1 plus t to the power of 24 end exponent right parenthesis$ is

$^{12} C_{6} + 2$
$^{12} C_{5}$
$^{12} C_{6}$
$^{12} C_{7}$

The correct answer is: $blank to the power of 12 end exponent C subscript 6 end subscript plus 2$

We have, $open parentheses 1 plus t to the power of 2 end exponent close parentheses to the power of 12 end exponent open parentheses 1 plus t to the power of 12 end exponent close parentheses left parenthesis 1 plus t to the power of 24 end exponent right parenthesis$
$equals open parentheses 1 plus blank to the power of 12 end exponent C subscript 1 end subscript t to the power of 2 end exponent plus blank to the power of 12 end exponent C subscript 2 end subscript t to the power of 4 end exponent plus... plus blank to the power of 12 end exponent C subscript 6 end subscript t to the power of 12 end exponent plus... plus blank to the power of 12 end exponent C subscript 12 end subscript t to the power of 24 end exponent plus... close parentheses left parenthesis 1 plus t to the power of 12 end exponent plus t to the power of 24 end exponent plus t to the power of 36 end exponent right parenthesis$
$therefore$ $C o e f f i c i e n t blank o f blank t to the power of 24 end exponent blank i n blank open parentheses 1 plus t to the power of 2 end exponent close parentheses to the power of 12 end exponent open parentheses 1 plus t to the power of 12 end exponent close parentheses left parenthesis 1 plus t to the power of 24 end exponent right parenthesis$
$equals blank to the power of 12 end exponent C subscript 6 end subscript plus blank to the power of 12 end exponent C subscript 12 end subscript plus 1 equals blank to the power of 12 end exponent C subscript 6 end subscript plus 2$

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