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

Question Number 11131 by Joel576 last updated on 13/Mar/17

Proof with induction, for every n \in N

\underset{i = 1}{\sum^{n}} 2^{i + 2} = 2^{n + 3} - 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

\Rightarrow 2^{3} = 2^{4} - 8

\Rightarrow 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} + \dots + n = 2^{n + 3} - 8

(2^{1 + 2} + 2^{2 + 2} + \dots + n) + (n + 1) = 2^{n + 3} - 8 + (n + 1)

2^{n + 4} - 8 = 2^{n + 3} - 8 + n + 1

Commented by mrW1 last updated on 13/Mar/17

i t s h o u l d b e :

S_{n} = 2^{1 + 2} + 2^{2 + 2} + \dots + 2^{n + 2} = 2^{n + 3} - 8

S_{n + 1} = (2^{1 + 2} + 2^{2 + 2} + \dots + 2^{n + 2}) + (2^{n + 1 + 2})

= 2^{n + 3} - 8 + 2^{n + 1 + 2}

= 2 \times 2^{n + 3} - 8

= 2^{(n + 1) + 3} - 8

Commented by Joel576 last updated on 13/Mar/17

o h, o k t h a n k y o u v e r y m u c h

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