If frac{(1-px)^{-1}}{(1-qx)}=a_0+a_1x+a_2x^2+ | vedclass

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

Question

(B) Given the expression: $\frac{(1-px)^{-1}}{(1-qx)} = a_0+a_1x+a_2x^2+\ldots+a_nx^n+\ldots$ Note that the expression is equivalent to $(1-px)^{-1}(1

Vedclass · Accepted Answer

$\frac{p^{n+1}-q^{n+1}}{p-q}$

Vedclass · Answer

(B) Given the expression: $\frac{(1-px)^{-1}}{(1-qx)} = a_0+a_1x+a_2x^2+\ldots+a_nx^n+\ldots$ Note that the expression is equivalent to $(1-px)^{-1}(1-qx)^{-1}$ only if the denominator was $(1-qx)^{-1}$. However,the problem states $\frac{(1-px)^{-1}}{(1-qx)}$. Wait,the standard expansion for $\frac{1}{(1-px)(1-qx)}$ is $\sum a_n x^n$. Using partial fractions: $\frac{1}{(1-px)(1-qx)} = \frac{A}{1-px} + \frac{B}{1-qx}$. $1 = A(1-qx) + B(1-px)$. For $x = 1/p$,$1 = A(1-q/p) \implies A = \frac{p}{p-q}$. For $x = 1/q$,$1 = B(1-p/q) \implies B = \frac{q}{q-p} = -\frac{q}{p-q}$. So,$\frac{1}{(1-px)(1-qx)} = \frac{p}{p-q} \sum (px)^n - \frac{q}{p-q} \sum (qx)^n$. $a_n = \frac{p}{p-q} p^n - \frac{q}{p-q} q^n = \frac{p^{n+1}-q^{n+1}}{p-q}$. Thus,$a_n = \frac{p^{n+1}-q^{n+1}}{p-q}$.

If $\frac{(1-px)^{-1}}{(1-qx)}=a_0+a_1x+a_2x^2+a_3x^3+\ldots$,then $a_n=$

Similar Questions

Which is larger: $(1.01)^{1000000}$ or $10,000$?

If $1 + x^4 + x^5 = \sum\limits_{i = 0}^5 a_i (1 + x)^i$ for all $x$ in $\mathbb{R},$ then $a_2$ is

The number of non-zero terms in the expansion of $(1 + 3\sqrt{2}x)^9 + (1 - 3\sqrt{2}x)^9$ is

If $\sum_{r=0}^{10} \left( \frac{10^{r+1}-1}{10^r} \right) \cdot {}^{11}C_{r+1} = \frac{\alpha^{11}-11^{11}}{10^{10}}$,then $\alpha$ is equal to :

The coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3) \ldots (1+x^{100})$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module