If the numerically greatest term in the expan | vedclass

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

Question

(A) For the numerically greatest term in the expansion of $(a+b)^n$,we consider the condition $T_{r+1} \geq T_r$.Given $(2-3x)^9$ with $x=1$,we have $

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(A) For the numerically greatest term in the expansion of $(a+b)^n$,we consider the condition $T_{r+1} \geq T_r$.Given $(2-3x)^9$ with $x=1$,we have $(2-3)^9 = (-1)^9$. We consider the absolute values: $|2-3|^9 = |-1|^9 = 1^9$.Let the expansion be $(2-3x)^9 = \sum_{r=0}^{9} {^9C_r} (2)^{9-r} (-3x)^r$.For $x=1$,the terms are $T_{r+1} = {^9C_r} 2^{9-r} (-3)^r$.The magnitude is $|T_{r+1}| = {^9C_r} 2^{9-r} 3^r$.We require $\frac{|T_{r+1}|}{|T_r|} \geq 1$.$\frac{{^9C_r} 2^{9-r} 3^r}{{^9C_{r-1}} 2^{9-(r-1)} 3^{r-1}} \geq 1 \Rightarrow \frac{9-r+1}{r} \cdot \frac{3}{2} \geq 1$.$3(10-r) \geq 2r$ $\Rightarrow 30-3r \geq 2r$ $\Rightarrow 5r \leq 30$ $\Rightarrow r \leq 6$.Thus,the greatest term occurs at $r=6$ and $r=7$ (since $r=6$ gives equality).For $r=6$,$|T_7| = {^9C_6} 2^3 3^6 = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} 	imes 2^3 	imes 3^6 = 84 	imes 8 	imes 729 = (2^2 	imes 3 	imes 7) 	imes 2^3 	imes 3^6 = 2^5 	imes 3^7 	imes 7^1$.The prime numbers are $P_1=2, P_2=3, P_3=7, P_4=11$ (as $P_4$ is not present,its exponent $\delta=0$).So,$\alpha=5, \beta=7, \gamma=1, \delta=0$.$\alpha+\beta+\gamma+\delta = 5+7+1+0 = 13$.

If the numerically greatest term in the expansion of $(2-3x)^9$ when $x=1$ is $P_1^\alpha P_2^\beta P_3^\gamma P_4^\delta$ (where $P_1 < P_2 < P_3 < P_4$ are the first four prime numbers),then $\alpha+\beta+\gamma+\delta=$

Similar Questions

The middle term in the expansion of ${\left( {1 - \frac{1}{x}} \right)^n}{\left( {1 - x} \right)^n}$ in powers of $x$ is

The coefficients of three consecutive terms of $(1+x)^{n+5}$ are in the ratio $5: 10: 14$. Then $n=$

If the sum of the coefficients in the expansion of $(x - 2y + 3z)^n$ is $128$,then the greatest coefficient in the expansion of $(1 + x)^n$ is

If $a_r$ is the coefficient of $x^{10-r}$ in the Binomial expansion of $(1+x)^{10}$,then $\sum \limits_{r=1}^{10} r^3\left(\frac{a_r}{a_{r-1}}\right)^2$ is equal to

Find the $4^{\text{th}}$ term in the expansion of $(x-2y)^{12}$. (in $x^9y^3$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module