The numerically greatest term in the expansio | vedclass

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(C) Given expansion is $(x+3y)^{13}$ with $x=\frac{1}{2}$ and $y=\frac{1}{3}$.Substituting the values,we get $(\frac{1}{2} + 3(\frac{1}{3}))^{13} = (\

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${ }^{13}C_9 \left(\frac{1}{2}\right)^9 \left(\frac{1}{3}\right)^4$

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(C) Given expansion is $(x+3y)^{13}$ with $x=\frac{1}{2}$ and $y=\frac{1}{3}$.Substituting the values,we get $(\frac{1}{2} + 3(\frac{1}{3}))^{13} = (\frac{1}{2} + 1)^{13} = (\frac{3}{2})^{13}$.Let $T_{r+1}$ be the $(r+1)$-th term in the expansion of $(a+b)^n$.The condition for the numerically greatest term is $\frac{T_{r+1}}{T_r} \geq 1$.Here,$T_{r+1} = {}^{13}C_r (\frac{1}{2})^{13-r} (1)^r = {}^{13}C_r (\frac{1}{2})^{13-r}$.$\frac{T_{r+1}}{T_r} = \frac{{}^{13}C_r (\frac{1}{2})^{13-r}}{{}^{13}C_{r-1} (\frac{1}{2})^{13-(r-1)}} = \frac{{}^{13}C_r}{{}^{13}C_{r-1}} 	imes 2 = \frac{13-r+1}{r} 	imes 2 = \frac{14-r}{r} 	imes 2$.Setting $\frac{28-2r}{r} \geq 1$,we get $28-2r \geq r$,so $3r \leq 28$,which means $r \leq 9.33$.Thus,$r=9$ gives the greatest term $T_{10} = {}^{13}C_9 (\frac{1}{2})^{13-9} (1)^9 = {}^{13}C_9 (\frac{1}{2})^4$.

The numerically greatest term in the expansion of $(x+3y)^{13}$,when $x=\frac{1}{2}$ and $y=\frac{1}{3}$ is

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