Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

$${oh},\:{ok}\:{thank}\:{you}\:{very}\:{much} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com