If the coefficients of (r-5)^{th} and (2r-1)^ | vedclass

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

Question

(A) The general term of the expansion $(1+x)^n$ is given by $T_{k+1} = {^nC_k} x^k$. The $(r-5)^{th}$ term is $T_{r-5} = T_{(r-6)+1}$,so its coefficie

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(A) The general term of the expansion $(1+x)^n$ is given by $T_{k+1} = {^nC_k} x^k$. The $(r-5)^{th}$ term is $T_{r-5} = T_{(r-6)+1}$,so its coefficient is ${^{34}C_{r-6}}$. The $(2r-1)^{th}$ term is $T_{2r-1} = T_{(2r-2)+1}$,so its coefficient is ${^{34}C_{2r-2}}$. Given that the coefficients are equal,we have ${^{34}C_{r-6}} = {^{34}C_{2r-2}}$. Using the property ${^nC_a} = {^nC_b} \implies a = b$ or $a + b = n$: Case $1$: $r-6 = 2r-2 \implies r = -4$ (Not possible as $r$ must be a natural number). Case $2$: $(r-6) + (2r-2) = 34 \implies 3r - 8 = 34 \implies 3r = 42 \implies r = 14$. Thus,$r = 14$.

If the coefficients of $(r-5)^{th}$ and $(2r-1)^{th}$ terms in the expansion of $(1+x)^{34}$ are equal,find $r$.

Similar Questions

If the coefficients of $x^7$ and $x^8$ in $(2 + \frac{x}{3})^n$ are equal,then $n$ is

If the coefficients of $x^{7}$ and $x^{8}$ in the expansion of $(2+\frac{x}{3})^{n}$ are equal,then the value of $n$ is equal to $.....$

The coefficient of $x^5$ in the expansion of $(1+x)^{21}+(1+x)^{22}+\ldots+(1+x)^{30}$ is

For each positive integer $n$,let $A_n = \max \left\{ \binom{n}{r} \mid 0 \leq r \leq n \right\}$. Then,the number of elements $n \in \{1, 2, \ldots, 20\}$ for which $1.9 \leq \frac{A_n}{A_{n-1}} \leq 2$ is

If the coefficients of $r$ th and $(r+1)$ th terms in the expansion of $(3+7x)^{29}$ are equal,then $r$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module