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Question

(A) For the expansion of $(a+b)^{n}$,if $n$ is even,the middle term is the $\left(\frac{n}{2}+1\right)^{th}$ term.Here,$n=10$,so the middle term is $\

Vedclass · Accepted Answer

$61236 x^{5} y^{5}$

Vedclass · Answer

(A) For the expansion of $(a+b)^{n}$,if $n$ is even,the middle term is the $\left(\frac{n}{2}+1ight)^{th}$ term.Here,$n=10$,so the middle term is $\left(\frac{10}{2}+1ight)^{th} = 6^{th}$ term.The general term $T_{r+1}$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r}$.For the $6^{th}$ term,$r=5$:$T_{6} = {}^{10}C_{5} \left(\frac{x}{3}ight)^{10-5} (9y)^{5}$$T_{6} = {}^{10}C_{5} \left(\frac{x}{3}ight)^{5} (9^{5} y^{5})$$T_{6} = \frac{10 	imes 9 	imes 8 	imes 7 	imes 6}{5 	imes 4 	imes 3 	imes 2 	imes 1} 	imes \frac{x^{5}}{3^{5}} 	imes (3^{2})^{5} y^{5}$$T_{6} = 252 	imes \frac{x^{5}}{3^{5}} 	imes 3^{10} y^{5}$$T_{6} = 252 	imes 3^{5} 	imes x^{5} y^{5}$$T_{6} = 252 	imes 243 	imes x^{5} y^{5} = 61236 x^{5} y^{5}$.

Find the middle term in the expansion of $\left(\frac{x}{3}+9 y\right)^{10}$.

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