For |x|

Question

(A) Given the expression $\frac{1}{(1-5x)(1-4x)}$ for $|x| < \frac{1}{5}$.Using the binomial expansion $(1-z)^{-1} = 1 + z + z^2 + z^3 + \dots$,we hav

Vedclass · Accepted Answer

$369$

Vedclass · Answer

(A) Given the expression $\frac{1}{(1-5x)(1-4x)}$ for $|x| Using the binomial expansion $(1-z)^{-1} = 1 + z + z^2 + z^3 + \dots$,we have:$(1-5x)^{-1} = 1 + 5x + 25x^2 + 125x^3 + \dots$$(1-4x)^{-1} = 1 + 4x + 16x^2 + 64x^3 + \dots$Multiplying these two series:$(1 + 5x + 25x^2 + 125x^3 + \dots)(1 + 4x + 16x^2 + 64x^3 + \dots)$The coefficient of $x^3$ is obtained by collecting terms that result in $x^3$:$1 \cdot (64x^3) + (5x) \cdot (16x^2) + (25x^2) \cdot (4x) + (125x^3) \cdot 1$$= 64 + 80 + 100 + 125$$= 369$

For $|x| < \frac{1}{5}$,the coefficient of $x^3$ in the expansion of $\frac{1}{(1-5 x)(1-4 x)}$ is

Similar Questions

The coefficient of $x^{4}$ in the expansion of $(1+x+x^{2})^{10}$ is

The first $3$ terms in the expansion of $(1 + ax)^n$ $(n \ne 0)$ are $1, 6x$ and $16x^2$. Then the value of $a$ and $n$ are respectively

$\binom{n}{0} + 2\binom{n}{1} + 2^2\binom{n}{2} + \dots + 2^n\binom{n}{n}$ is equal to

Using the Binomial Theorem,evaluate $(99)^{5}$.

If $(1 + x - 2x^2)^6 = 1 + a_1x + a_2x^2 + .... + a_{12}x^{12}$,then the expression $a_2 + a_4 + a_6 + .... + a_{12}$ has the value

Vedclass Test Series

Exam Paper Generator

Online Exam Module