Coefficient of t^{20} in the expansion of (1 | vedclass

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Question

$\left(1+\mathrm{t}^{2}\right)^{10}\left(1+\mathrm{t}^{10}+\mathrm{t}^{20}+\mathrm{t}^{30}\right)$

$ = \left( {1 + {\,^{10}}{{\rm{C}}_1}{{\rm{t}}^2

Vedclass · Accepted Answer

$^{10}C_5 + 2$

Vedclass · Answer

$\left(1+\mathrm{t}^{2}ight)^{10}\left(1+\mathrm{t}^{10}+\mathrm{t}^{20}+\mathrm{t}^{30}ight)$

$ = \left( {1 + {\,^{10}}{{m{C}}_1}{{m{t}}^2} + {\,^{10}}{{m{C}}_2}{{m{t}}^4} +  \ldots . + {\,^{10}}{{m{C}}_{10}}{{m{t}}^{20}}} ight)$

$\left( {1 + {t^{10}} + {t^{20}} + {t^{30}}} ight)$

$	herefore $ Coefficient $ = {\,^{10}}{{m{C}}_{10}} + {\,^{10}}{{m{C}}_5} + {\,^{10}}{{m{C}}_0} = 2 + {\,^{10}}{{m{C}}_5}$

Coefficient of $t^{20}$ in the expansion of $(1 + t^2)^{10}(1 + t^{10})(1 + t^{20})$ is

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