Question Number 29970 by abdo imad last updated on 14/Feb/18
$${a}>\mathrm{0}\:{and}\:{b}>\mathrm{0}\:\:{if}\:\:\:\frac{\mathrm{1}}{\left(\mathrm{1}−{ax}\right)\left(\mathrm{1}−{bx}\right)}=\sum_{{n}} \:\:{a}_{{n}} \:{x}^{{n}} \\ $$$${find}\:{the}\:{sequence}\:{a}_{{n}} . \\ $$
Commented by abdo imad last updated on 16/Feb/18
$${for}\:\mid{x}\mid<\:{inf}\left(\frac{\mathrm{1}}{\mid{a}\mid},\frac{\mathrm{1}}{\mid{b}\mid}\right)\:{we}\:{have} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−{ax}\right)\left(\mathrm{1}−{bx}\right)}\:=\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}^{{n}} {x}^{{n}} \right)\left(\sum_{{m}=\mathrm{0}} ^{\infty} \:{b}^{{m}} \:{x}^{{m}} \right) \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\infty} \:\:{c}_{{k}\:} \:{x}^{{k}} \:\:{with}\:\:{c}_{{k}} =\:\sum_{{i}+{j}={k}} {u}_{{i}} \:{v}_{{j}} =\:\sum_{{i}+{j}={k}} \:{a}^{{i}} \:{b}^{{j}} \\ $$$$=\sum_{{i}=\mathrm{0}} ^{{k}} \:{a}^{{i}} \:{b}^{{k}−{i}} =\frac{{a}^{{k}+\mathrm{1}} \:−{b}^{{k}+\mathrm{1}} }{{a}−{b}}\:{if}\:{a}\neq{b}\:{and}\:{if}\:{a}={b} \\ $$$${c}_{{k}} =\sum_{{i}=\mathrm{0}} ^{{k}} \:{a}^{{i}} =\frac{\mathrm{1}−{a}^{{k}+\mathrm{1}} }{\mathrm{1}−{a}}\:{if}\:{a}\neq\mathrm{1}\:{and}\:{c}_{{k}} =\left({k}+\mathrm{1}\right)\:{if}\:{a}=\mathrm{1}{finally} \\ $$$${if}\:{a}\neq{b}\:\:{a}_{{n}} =\frac{{a}^{{n}+\mathrm{1}} \:−{b}^{{n}+\mathrm{1}} }{{a}−{b}}\:{anf}\:{if}\:{a}={b}\:\neq\mathrm{1}\:{a}_{{n}} =\frac{\mathrm{1}−{a}^{{n}+\mathrm{1}} }{\mathrm{1}−{a}} \\ $$$${if}\:{a}={b}=\mathrm{1}\:\:\:{a}_{{n}} ={n}+\mathrm{1}\:{and}\:{we}\:{get}\:{the}\:{formula} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left({n}+\mathrm{1}\right){x}^{{n}} . \\ $$$$ \\ $$