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વિધેય $\frac{1}{{\left( {1 - ax} \right)\left( {1 - bx} \right)}}$ નુ $x$ ની ધાતાકમાં વિસ્તરણ ${a_0} + {a_1}x + {a_2}{x^2} + \;{a_3}{x^3} + \; \ldots......$ હોય તો ${a_n}$ મેળવો.
$\frac{{{b^n} - {a^n}}}{{b - a}}$
$\;\frac{{{a^n} - {b^n}}}{{b - a}}$
$\;\frac{{{a^{n + 1}} - {b^{n + 1}}}}{{b - a}}$
$\;\frac{{{b^{n + 1}} - {a^{n + 1}}}}{{b - a}}$
Solution
$\frac{1}{(1-a x)(1-b x)}$
$=(1-a x)^{-1}(1-b x)^{-1}$
$=\left[1+(-1)(-a x)+\frac{(-1)(-2)}{1.2}(-a x)^{2}+\ldots\right]$$\left[1+(-1)(-b x)+\frac{(-1)(-2)}{1.2}(-b x)^{2}+\ldots\right]$
$=\left[1+a x+a^{2} x^{2}+\ldots+a^{n-1} x^{n-1}+a^{n} x^{n}+\ldots .\right]$$\left[1+b x+b^{2} x^{2}+\ldots+b^{n-1} x^{n-1}+b^{n} x^{n}+\ldots\right]$
Coefficient of $x^{n}=$
$a^{n}+a^{n-1} b+a^{n-2} b^{2}+\ldots .+b^{n}$
$=a^{n}\left[1+\frac{b}{a}+\frac{b^{2}}{a^{2}}+\ldots \ldots+\frac{b^{n}}{a^{n}}\right]$
$=a^{n}\left[\frac{\left(\frac{b}{a}\right)^{n+1}-1}{\frac{b}{a}-1}\right]$
$=a^{n}\left[\frac{b^{n+1}-a^{n+1}}{a^{n+1}\left(\frac{b-a}{a}\right)}\right]$
$=\frac{b^{n+1}-a^{n+1}}{b-a}$
