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Question Number 133146 by shaker last updated on 19/Feb/21

Commented by Dwaipayan Shikari last updated on 19/Feb/21

((1−a^(n+1) )/(1−a))

$$\frac{\mathrm{1}−{a}^{{n}+\mathrm{1}} }{\mathrm{1}−{a}}\:\:\:\: \\ $$

Answered by mathmax by abdo last updated on 19/Feb/21

if a≠1  we have 1+a+a^2 +...+a^n  =((1−a^(n+1) )/(1−a))  if a=1  we get 1+a+a^2 +...+a^n  =n+1

$$\mathrm{if}\:\mathrm{a}\neq\mathrm{1}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{1}+\mathrm{a}+\mathrm{a}^{\mathrm{2}} +...+\mathrm{a}^{\mathrm{n}} \:=\frac{\mathrm{1}−\mathrm{a}^{\mathrm{n}+\mathrm{1}} }{\mathrm{1}−\mathrm{a}} \\ $$$$\mathrm{if}\:\mathrm{a}=\mathrm{1}\:\:\mathrm{we}\:\mathrm{get}\:\mathrm{1}+\mathrm{a}+\mathrm{a}^{\mathrm{2}} +...+\mathrm{a}^{\mathrm{n}} \:=\mathrm{n}+\mathrm{1} \\ $$

Answered by physicstutes last updated on 19/Feb/21

a = 1 , r = a    S_n  = ((1(1−a^(n+1) ))/(1−a))

$${a}\:=\:\mathrm{1}\:,\:{r}\:=\:{a} \\ $$$$\:\:\mathrm{S}_{{n}} \:=\:\frac{\mathrm{1}\left(\mathrm{1}−{a}^{{n}+\mathrm{1}} \right)}{\mathrm{1}−{a}} \\ $$

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