Find the coefficients of x^r where 0 le r le | vedclass

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

Question

(B) The given expression is a geometric series with first term $a = (x + 3)^{n - 1}$,common ratio $q = \frac{x + 2}{x + 3}$,and $n$ terms.
Using the

Vedclass · Accepted Answer

$^nC_r(3^{n - r} - 2^{n - r})$

Vedclass · Answer

(B) The given expression is a geometric series with first term $a = (x + 3)^{n - 1}$,common ratio $q = \frac{x + 2}{x + 3}$,and $n$ terms.
Using the sum formula for a geometric series $S_n = \frac{a(1 - q^n)}{1 - q}$:
$S_n = \frac{(x + 3)^{n - 1} [1 - (\frac{x + 2}{x + 3})^n]}{1 - \frac{x + 2}{x + 3}} = \frac{(x + 3)^{n - 1} [\frac{(x + 3)^n - (x + 2)^n}{(x + 3)^n}]}{\frac{(x + 3) - (x + 2)}{x + 3}} = \frac{(x + 3)^n - (x + 2)^n}{(x + 3) - (x + 2)} = (x + 3)^n - (x + 2)^n$.
Now,we find the coefficient of $x^r$ in the expansion of $(x + 3)^n - (x + 2)^n$.
Using the Binomial Theorem,$(x + a)^n = \sum_{k=0}^{n} {^nC_k} x^{n-k} a^k$.
For $(x + 3)^n$,the term containing $x^r$ is $^nC_{n-r} x^r 3^{n-r} = ^nC_r x^r 3^{n-r}$.
For $(x + 2)^n$,the term containing $x^r$ is $^nC_{n-r} x^r 2^{n-r} = ^nC_r x^r 2^{n-r}$.
Thus,the coefficient of $x^r$ is $^nC_r 3^{n-r} - ^nC_r 2^{n-r} = ^nC_r(3^{n - r} - 2^{n - r})$.

Find the coefficients of $x^r$ where $0 \le r \le (n - 1)$ in the expansion of $(x + 3)^{n - 1} + (x + 3)^{n - 2}(x + 2) + (x + 3)^{n - 3}(x + 2)^2 + ... + (x + 2)^{n - 1}$.

Similar Questions

The value of $x$ in the expression $(x + x^{\log_{10} x})^5$,if the third term in the expansion is $1,000,000$.

In the expansion of ${\left( \frac{a}{x} + bx \right)^{12}}$,the coefficient of $x^{-10}$ is:

If $x + y = 1$,then $\sum\limits_{r = 0}^n {{r^2}{\,^n}{C_r}{x^r}{y^{n - r}}} $ equals

The numerically greatest term in the expansion of $(3x - 16y)^{15}$ when $x = \frac{2}{3}$ and $y = \frac{3}{2}$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module