The 13^{text{th}} term in the expansion of (1 | vedclass

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(C) The general term $T_{r+1}$ in the expansion of $(1-ax)^{-n}$ is given by ${}^{n+r-1}C_r (ax)^r$. Here,$n=4$,$a=4$,and we need the $13^{	ext{th}}$

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${}^{15}C_3 4^{12} x^{12}$

Vedclass · Answer

(C) The general term $T_{r+1}$ in the expansion of $(1-ax)^{-n}$ is given by ${}^{n+r-1}C_r (ax)^r$. Here,$n=4$,$a=4$,and we need the $13^{	ext{th}}$ term,so $r+1=13$,which means $r=12$. Substituting these values,$T_{13} = {}^{4+12-1}C_{12} (4x)^{12}$. $T_{13} = {}^{15}C_{12} (4x)^{12}$. Using the property ${}^nC_r = {}^nC_{n-r}$,we have ${}^{15}C_{12} = {}^{15}C_{15-12} = {}^{15}C_3$. Therefore,$T_{13} = {}^{15}C_3 4^{12} x^{12}$.

The $13^{\text{th}}$ term in the expansion of $(1-4x)^{-4}$ is

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