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Proof-with-induction-for-every-n-N-i-1-n-2-i-2-2-n-3-8-

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Question Number 11131 by Joel576 last updated on 13/Mar/17
Proof with induction, for every n ∈ N Σ_(i = 1) ^n 2^(i + 2) = 2^(n + 3) − 8
$$\mathrm{Proof}\:\mathrm{with}\:\mathrm{induction},\:\mathrm{for}\:\mathrm{every}\:{n}\:\in\:\mathbb{N} \\ $$$$\underset{{i}\:=\:\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{2}^{{i}\:+\:\mathrm{2}} \:=\:\mathrm{2}^{{n}\:+\:\mathrm{3}} \:−\:\mathrm{8} \\ $$
Commented by Joel576 last updated on 13/Mar/17
I have answered it, but it seems I have done something wrong • For n = 1 ⇒ 2^(3 ) = 2^4 − 8 ⇒ 8 = 8 (proved) • If it proved for n, so it also proved for n + 1 S_n = 2^(n + 3) − 8 S_(n + 1) = 2^(n + 4) − 8 2^(1 + 2) + 2^(2 + 2) + ... + n = 2^(n + 3) − 8 (2^(1 + 2) + 2^(2 + 2) + ... + n) + (n + 1) = 2^(n + 3) − 8 + (n + 1) 2^(n + 4) − 8 = 2^(n + 3) − 8 + n + 1
$$\mathrm{I}\:\mathrm{have}\:\mathrm{answered}\:\mathrm{it},\:\mathrm{but}\:\mathrm{it}\:\mathrm{seems}\:\mathrm{I}\:\mathrm{have}\:\mathrm{done}\:\mathrm{something}\:\mathrm{wrong} \\ $$$$ \\ $$$$\bullet\:\mathrm{For}\:{n}\:=\:\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{2}^{\mathrm{3}\:} \:=\:\mathrm{2}^{\mathrm{4}} \:−\:\mathrm{8} \\ $$$$\Rightarrow\:\mathrm{8}\:\:\:=\:\mathrm{8}\:\left(\mathrm{proved}\right) \\ $$$$ \\ $$$$\bullet\:\mathrm{If}\:\mathrm{it}\:\mathrm{proved}\:\mathrm{for}\:{n},\:\mathrm{so}\:\mathrm{it}\:\mathrm{also}\:\mathrm{proved}\:\mathrm{for}\:{n}\:+\:\mathrm{1} \\ $$$${S}_{{n}} \:=\:\mathrm{2}^{{n}\:+\:\mathrm{3}} \:−\:\mathrm{8} \\ $$$${S}_{{n}\:+\:\mathrm{1}} \:=\:\mathrm{2}^{{n}\:+\:\mathrm{4}} \:−\:\mathrm{8} \\ $$$$ \\ $$$$\mathrm{2}^{\mathrm{1}\:+\:\mathrm{2}} \:+\:\mathrm{2}^{\mathrm{2}\:+\:\mathrm{2}} \:+\:…\:+\:{n}\:=\:\mathrm{2}^{{n}\:+\:\mathrm{3}} \:−\:\mathrm{8} \\ $$$$\left(\mathrm{2}^{\mathrm{1}\:+\:\mathrm{2}} \:+\:\mathrm{2}^{\mathrm{2}\:+\:\mathrm{2}} \:+\:…\:+\:{n}\right)\:+\:\left({n}\:+\:\mathrm{1}\right)\:=\:\mathrm{2}^{{n}\:+\:\mathrm{3}} \:−\:\mathrm{8}\:+\:\left({n}\:+\:\mathrm{1}\right) \\ $$$$\mathrm{2}^{{n}\:+\:\mathrm{4}} \:−\:\mathrm{8}\:=\:\mathrm{2}^{{n}\:+\:\mathrm{3}} \:−\:\mathrm{8}\:+\:{n}\:+\:\mathrm{1} \\ $$
Commented by mrW1 last updated on 13/Mar/17
it should be: S_n =2^(1 + 2) + 2^(2 + 2) + ... + 2^(n+2) = 2^(n + 3) − 8 S_(n+1) =(2^(1 + 2) + 2^(2 + 2) + ... + 2^(n+2) ) + (2^(n+1+2) ) =2^(n + 3) − 8+2^(n+1+2) =2×2^(n+3) −8 =2^((n+1)+3) −8
$${it}\:{should}\:{be}: \\ $$$${S}_{{n}} =\mathrm{2}^{\mathrm{1}\:+\:\mathrm{2}} \:+\:\mathrm{2}^{\mathrm{2}\:+\:\mathrm{2}} \:+\:…\:+\:\mathrm{2}^{{n}+\mathrm{2}} \:=\:\mathrm{2}^{{n}\:+\:\mathrm{3}} \:−\:\mathrm{8} \\ $$$${S}_{{n}+\mathrm{1}} =\left(\mathrm{2}^{\mathrm{1}\:+\:\mathrm{2}} \:+\:\mathrm{2}^{\mathrm{2}\:+\:\mathrm{2}} \:+\:…\:+\:\mathrm{2}^{{n}+\mathrm{2}} \right)\:+\:\left(\mathrm{2}^{{n}+\mathrm{1}+\mathrm{2}} \right)\: \\ $$$$=\mathrm{2}^{{n}\:+\:\mathrm{3}} \:−\:\mathrm{8}+\mathrm{2}^{{n}+\mathrm{1}+\mathrm{2}} \\ $$$$=\mathrm{2}×\mathrm{2}^{{n}+\mathrm{3}} −\mathrm{8} \\ $$$$=\mathrm{2}^{\left({n}+\mathrm{1}\right)+\mathrm{3}} −\mathrm{8} \\ $$
Commented by Joel576 last updated on 13/Mar/17
oh, ok thank you very much
$${oh},\:{ok}\:{thank}\:{you}\:{very}\:{much} \\ $$

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