Question Number 133146 by shaker last updated on 19/Feb/21
Commented by Dwaipayan Shikari last updated on 19/Feb/21
$$\frac{\mathrm{1}−{a}^{{n}+\mathrm{1}} }{\mathrm{1}−{a}}\:\:\:\: \\ $$
Answered by mathmax by abdo last updated on 19/Feb/21
$$\mathrm{if}\:\mathrm{a}\neq\mathrm{1}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{1}+\mathrm{a}+\mathrm{a}^{\mathrm{2}} +...+\mathrm{a}^{\mathrm{n}} \:=\frac{\mathrm{1}−\mathrm{a}^{\mathrm{n}+\mathrm{1}} }{\mathrm{1}−\mathrm{a}} \\ $$$$\mathrm{if}\:\mathrm{a}=\mathrm{1}\:\:\mathrm{we}\:\mathrm{get}\:\mathrm{1}+\mathrm{a}+\mathrm{a}^{\mathrm{2}} +...+\mathrm{a}^{\mathrm{n}} \:=\mathrm{n}+\mathrm{1} \\ $$
Answered by physicstutes last updated on 19/Feb/21
$${a}\:=\:\mathrm{1}\:,\:{r}\:=\:{a} \\ $$$$\:\:\mathrm{S}_{{n}} \:=\:\frac{\mathrm{1}\left(\mathrm{1}−{a}^{{n}+\mathrm{1}} \right)}{\mathrm{1}−{a}} \\ $$