The numerically greatest term in the expansio | vedclass

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Question

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

Vedclass · Answer

(A) 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 $(\frac{1}{2} + 1)^{13}$.$T_{r+1} = {}^{13}C_r (\frac{1}{2})^{13-r} (1)^r = {}^{13}C_r (\frac{1}{2})^{13-r}$.The condition for the numerically greatest term is $\frac{T_{r+1}}{T_r} \geq 1$.$\frac{T_{r+1}}{T_r} = \frac{{}^{13}C_r (\frac{1}{2})^{13-r}}{{}^{13}C_{r-1} (\frac{1}{2})^{13-r+1}} = \frac{{}^{13}C_r}{{}^{13}C_{r-1}} 	imes 2 = \frac{13-r+1}{r} 	imes 2 = \frac{28-2r}{r}$.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 $T_{10} = {}^{13}C_9 (\frac{1}{2})^{13-9} = {}^{13}C_9 (\frac{1}{2})^4$.

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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