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Question Number 29970 by abdo imad last updated on 14/Feb/18

$$a>0andb>0if\frac{1}{(1-ax)(1-bx)}=\sum _n{a}_n{x}^n$$ $$findthesequence{a}_n.$$

Commented byabdo imad last updated on 16/Feb/18

$$for\mid x\mid <inf(\frac{1}{\mid a\mid },\frac{1}{\mid b\mid })wehave$$ $$\frac{1}{(1-ax)(1-bx)}=\left(\sum _{n=0}^{\mathrm{\infty }}{a}^n{x}^n\right)\left(\sum _{m=0}^{\mathrm{\infty }}{b}^m{x}^m\right)$$ $$=\sum _{k=0}^{\mathrm{\infty }}{c}_k{x}^{k}with{c}_k=\sum _{i+j=k}{u}_i{v}_j=\sum _{i+j=k}{a}^i{b}^j$$ $$=\sum _{i=0}^k{a}^i{b}^{k-i}=\frac{a^{k+1}-b^{k+1}}{a-b}ifa\ne bandifa=b$$ $$c_k=\sum _{i=0}^k{a}^i=\frac{1-a^{k+1}}{1-a}ifa\ne 1and{c}_k=(k+1)ifa=1finally$$ $$ifa\ne b{a}_n=\frac{a^{n+1}-b^{n+1}}{a-b}anfifa=b\ne 1a_n=\frac{1-a^{n+1}}{1-a}$$ $$ifa=b=1a_n=n+1andwegettheformula$$ $$\frac{1}{{(1-x)}^2}=\sum _{n=0}^{\mathrm{\infty }}(n+1)x^n.$$ $$$$

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