The r^{th} term in the expansion of (a + 2x)^ | vedclass

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

Question

(B) The general term formula for the expansion of $(a + b)^n$ is $T_{r+1} = ^nC_r a^{n-r} b^r$.To find the $r^{th}$ term,we substitute $r$ with $r-1$

Vedclass · Accepted Answer

$\frac{n(n - 1)...(n - r + 2)}{(r - 1)!} a^{n - r + 1} (2x)^{r - 1}$

Vedclass · Answer

(B) The general term formula for the expansion of $(a + b)^n$ is $T_{r+1} = ^nC_r a^{n-r} b^r$.To find the $r^{th}$ term,we substitute $r$ with $r-1$ in the general term formula:$T_r = T_{(r-1)+1} = ^nC_{r-1} a^{n-(r-1)} (2x)^{r-1}$$T_r = ^nC_{r-1} a^{n-r+1} (2x)^{r-1}$Using the definition of combinations,$^nC_{r-1} = \frac{n!}{(n-r+1)!(r-1)!} = \frac{n(n-1)...(n-r+2)}{(r-1)!}$.Thus,$T_r = \frac{n(n-1)...(n-r+2)}{(r-1)!} a^{n-r+1} (2x)^{r-1}$.

The $r^{th}$ term in the expansion of $(a + 2x)^n$ is

Similar Questions

In the binomial expansion of $(a-b)^n, n \geq 5$,the sum of the $5^{\text{th}}$ and $6^{\text{th}}$ terms is zero. Then the value of $\frac{a}{b}$ is

In the expansion of ${\left( {3x - \frac{1}{{{x^2}}}} \right)^{10}}$,the $5^{th}$ term from the end is:

Given that the $4^{th}$ term in the expansion of $(2 + \frac{3}{8}x)^{10}$ has the maximum numerical value,the range of values of $x$ for which this will be true is given by:

The number of rational terms in the binomial expansion of $(\sqrt[4]{5}+\sqrt[5]{4})^{100}$ is

The coefficients of three consecutive terms in the expansion of $(1+a)^{n}$ are in the ratio $1:7:42$. Find $n$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module