In the binomial expansion of (a - b)^n, n ge | vedclass

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

Question

(B) The general term in the expansion of $(a - b)^n$ is given by $T_{r+1} = ^nC_r a^{n-r} (-b)^r$.For the $5^{th}$ term $(r=4)$: $T_5 = ^nC_4 a^{n-4}

Vedclass · Accepted Answer

$\frac{n - 4}{5}$

Vedclass · Answer

(B) The general term in the expansion of $(a - b)^n$ is given by $T_{r+1} = ^nC_r a^{n-r} (-b)^r$.For the $5^{th}$ term $(r=4)$: $T_5 = ^nC_4 a^{n-4} (-b)^4 = ^nC_4 a^{n-4} b^4$.For the $6^{th}$ term $(r=5)$: $T_6 = ^nC_5 a^{n-5} (-b)^5 = -^nC_5 a^{n-5} b^5$.Given $T_5 + T_6 = 0$,we have:$^nC_4 a^{n-4} b^4 - ^nC_5 a^{n-5} b^5 = 0$.Rearranging the terms:$^nC_4 a^{n-4} b^4 = ^nC_5 a^{n-5} b^5$.Dividing both sides by $a^{n-5} b^4$:$\frac{a}{b} = \frac{^nC_5}{^nC_4}$.Using the formula $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$:$\frac{a}{b} = \frac{n-5+1}{5} = \frac{n-4}{5}$.

In the binomial expansion of $(a - b)^n, n \ge 5,$ the sum of the $5^{th}$ and $6^{th}$ terms is zero. Then $a/b$ equals:

Similar Questions

Find the coefficient of $x^{15}$ in the product $(1 - x) (1 - 2x) (1 - 2^2 x) (1 - 2^3 x) \dots (1 - 2^{15} x)$.

If $p$ and $q$ are real numbers such that the $7^{\text{th}}$ term in the expansion of $\left(\frac{5}{p^3} - \frac{3q}{7}\right)^8$ is $700$,then $49p^2 =$ (in $q^2$)

The greatest term in the expansion of $(1+x)^{15}$,when $x=\frac{1}{2}$ is

The numerically greatest term in the binomial expansion of $(2x - 3y)^5$ when $x = \frac{3}{2}$ and $y = \frac{2}{3}$ is

The term independent of $x$ in the expansion ${\left( {{x^2} - \frac{1}{{3x}}} \right)^9}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module