Jf {(1-px)^{-1}}{1-qx} = a₀ + a₁ x + a₂ x² + a₃ x³ + ... , then aₙ =

Correct Answer is : pn+1−qn+1p−q{pⁿ+1}-qⁿ+1}}{p-q}p−qpn+1−qn+1

Correct Answer is : pn+1−qn+1p−q{pⁿ+1}-qⁿ+1}}{p-q}p−qpn+1−qn+1

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AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 18 of 160

Marks: +1, -0

\frac{(1-px)^{-1}}{1-qx}

a_0 + a_1 x + a_2 x^2 + a_3 x^3

a_n

Solution:

Consider the expression,

\frac{(1-p x)^{-1}}{1-q x}=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+\dots

(1-p x)^{-1}=1+p x+p^{2} x^{2}+p^{3} x^{3}+\dots+p^{n} x^{n}+\dots

And,

(1-q x)^{-1}=1+q x+q^{2} x^{2}+q^{3} x^{3}+\dots+q^{n} x^{n}+\dots

Then coefficient of

x^{n}

in the expansion of

(1-p x)^{-1}(1-q x)^{-1}=p^{n}+p^{n-1} q+p^{n-2} q^{2}+p^{n-3} q^{3}+\dots+q^{n}

Then,

a_n = \frac{p^{n}\left(1-\left(\frac{q}{p}\right)^{n+1}\right)}{1-\frac{q}{p}}

= \frac{p^{n}(p^{n+1}-q^{n+1}) p}{(p-q) p^{n+1}}

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

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