Find the 13^{text{th}} term in the expansion | vedclass

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(A) The $(r+1)^{	ext{th}}$ term in the binomial expansion of $(a+b)^n$ is given by $T_{r+1} = {^nC_r} a^{n-r} b^r$.For the expansion of $\left(9x - \

Vedclass · Accepted Answer

$18564$

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(A) The $(r+1)^{	ext{th}}$ term in the binomial expansion of $(a+b)^n$ is given by $T_{r+1} = {^nC_r} a^{n-r} b^r$.For the expansion of $\left(9x - \frac{1}{3\sqrt{x}}ight)^{18}$,we have $n=18$,$a=9x$,and $b=-\frac{1}{3\sqrt{x}}$.To find the $13^{	ext{th}}$ term,we set $r+1 = 13$,which gives $r=12$.$T_{13} = {^{18}C_{12}} (9x)^{18-12} \left(-\frac{1}{3\sqrt{x}}ight)^{12}$$T_{13} = {^{18}C_{12}} (9x)^6 \left(\frac{1}{3\sqrt{x}}ight)^{12}$$T_{13} = \frac{18!}{12!6!} \cdot (3^2)^6 \cdot x^6 \cdot \frac{1}{3^{12} \cdot (x^{1/2})^{12}}$$T_{13} = \frac{18 	imes 17 	imes 16 	imes 15 	imes 14 	imes 13}{6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1} \cdot 3^{12} \cdot x^6 \cdot \frac{1}{3^{12} \cdot x^6}$$T_{13} = 18564 \cdot 1 \cdot 1 = 18564$.

Find the $13^{\text{th}}$ term in the expansion of $\left(9x - \frac{1}{3\sqrt{x}}\right)^{18}, x \neq 0$.

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