Coefficient of t^{50} in (1 + t^2)^{25}(1 + t^{25})(1 + t^{40})(1 + t^{45})(1 + t^{47}) is -

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7.Binomial Theorem

normal

The coefficient of $t^{50}$ in $(1 + t^2)^{25}(1 + t^{25})(1 + t^{40})(1 + t^{45})(1 + t^{47})$ is -

$1 + ^{25}C_5$

$1 + ^{25}C_5 + ^{25}C_7$

$1 + ^{25}C_7$

$2 + ^{25}C_5$

$\left(1+t^{2}\right)^{25}\left(1+t^{25}+t^{40}+t^{45}+t^{47}+\ldots . .\right)$

${\,^{25}}{{\rm{C}}_{25}}{\left( {{{\rm{t}}^2}} \right)^{25}} \cdot 1 + {\,^{25}}{{\rm{C}}_5}{\left( {{{\rm{t}}^2}} \right)^5} \cdot {{\rm{t}}^{40}}$

$=$ coefficient of ${t^{50}} = {\,^{25}}{C_{25}} + {\,^{25}}{C_5} = 1 + {\,^{25}}{C_5}$

Standard 11

Mathematics

hard

hard

(JEE MAIN-2014)

easy

hard

(JEE MAIN-2014)

medium