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

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