Question Number 68151 by mhmd last updated on 06/Sep/19
$$\mathrm{If}\:\mid{a}\mid<\:\mathrm{1}\:\mathrm{and}\:\mid{b}\mid<\:\mathrm{1},\:\mathrm{then}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series} \\ $$ $$\mathrm{1}+\left(\mathrm{1}+{a}\right){b}+\left(\mathrm{1}+{a}+{a}^{\mathrm{2}} \right){b}^{\mathrm{2}} +\left(\mathrm{1}+{a}+{a}^{\mathrm{2}} +{a}^{\mathrm{3}} \right){b}^{\mathrm{3}} +... \\ $$ $$\mathrm{is} \\ $$
Commented by~ À ® @ 237 ~ last updated on 06/Sep/19
$${Let}\:{it}\:{be}\:\:{S}\:.\:{We}\:{can}\:{ascertain}\:{that} \\ $$ $${S}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{1}+{a}+{a}^{\mathrm{2}} +.....+{a}^{{n}} \right){b}^{{n}} \: \\ $$ $${knowing}\:{that}\:\:\mathrm{1}+{a}+...+{a}^{{n}} =\:\frac{\mathrm{1}−{a}^{{n}+\mathrm{1}} }{\mathrm{1}−{a}}\: \\ $$ $${S}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\left(\frac{\mathrm{1}−{a}^{{n}+\mathrm{1}} }{\mathrm{1}−{a}}\right){b}^{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{1}−{a}}\:\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:{b}^{{n}} \:−\:{a}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left({ab}\right)^{{n}} \right) \\ $$ $${Using}\:\:\frac{\mathrm{1}}{\mathrm{1}−{t}}=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:{t}^{{n}} \: \\ $$ $${S}=\frac{\mathrm{1}}{\mathrm{1}−{a}}\left(\frac{\mathrm{1}}{\mathrm{1}−{b}}−\frac{{a}}{\mathrm{1}−{ab}}\right)\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{b}\right)\left(\mathrm{1}−{ab}\right)}\: \\ $$