The numerically greatest term in the expansio | vedclass

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(B) Let $T_{r+1}$ be the greatest term,so $\frac{T_{r+1}}{T_r} \geq 1$. For the expansion $(a+b)^n$,the condition is $\frac{T_{r+1}}{T_r} = \frac{n-r+

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$3^3 	imes 5^5$

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(B) Let $T_{r+1}$ be the greatest term,so $\frac{T_{r+1}}{T_r} \geq 1$. For the expansion $(a+b)^n$,the condition is $\frac{T_{r+1}}{T_r} = \frac{n-r+1}{r} 	imes |\frac{b}{a}| \geq 1$. Here $n=6$,$a=5$,$b=3x$. At $x=1$,$b=3$. $\frac{6-r+1}{r} 	imes \frac{3}{5} \geq 1$ $\Rightarrow \frac{7-r}{r} 	imes \frac{3}{5} \geq 1$ $\Rightarrow 21 - 3r \geq 5r$ $\Rightarrow 8r \leq 21$ $\Rightarrow r \leq 2.625$. Since $r$ must be an integer,the greatest term occurs at $r=2$,which is $T_{2+1} = T_3$. $T_3 = {}^6C_2 	imes 5^{6-2} 	imes (3 	imes 1)^2$ $T_3 = 15 	imes 5^4 	imes 3^2$ $T_3 = (3 	imes 5) 	imes 5^4 	imes 3^2 = 3^3 	imes 5^5$.

The numerically greatest term in the expansion of $(5+3x)^6$ when $x=1$ is:

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