The numerically greatest term in the expansio | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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+

Vedclass · Accepted Answer

$3^3 	imes 5^5$

Vedclass · Answer

(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:

Similar Questions

The sum of the binomial coefficients of $[2x + \frac{1}{x}]^n$ is equal to $256$. The constant term in the expansion is

The coefficient of $x^2$ in the expansion of $(2x^2 + \frac{1}{x})^{10}, x \neq 0$, is:

Find the term independent of $x$ $(x > 0, x \neq 1)$ in the expansion of $\left[\frac{(x+1)}{\left(x^{2/3}-x^{1/3}+1\right)}-\frac{(x-1)}{(x-\sqrt{x})}\right]^{10}$.

The coefficient of $x^{91}$ in the series $^{100}C_1 \cdot 2^8 \cdot (1 - x)^{99} + ^{100}C_2 \cdot 2^7 \cdot (1 - x)^{98} + ^{100}C_3 \cdot 2^6 \cdot (1 - x)^{97} + \dots + ^{100}C_9 \cdot (1 - x)^{91}$ is equal to -

If the constant term in the expansion of $\left(\frac{\sqrt[5]{3}}{x}+\frac{2x}{\sqrt[3]{5}}\right)^{12}, x \neq 0$,is $\alpha \times 2^8 \times \sqrt[5]{3}$,then $25 \alpha$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module