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Re: on 2^0 + 2^{1} + ... + 2^{n} <= 2 for all n
Posted:
Apr 9, 2016 3:35 PM


On Saturday, April 9, 2016 at 12:52:09 PM UTC5, Daniel Bastos wrote: > Prove by induction > > 2^0 + 2^{1} + 2^{2} + 2^{3} + ... + 2^{n} <= 2 > > for all natural n. > > I haven't found a proof by induction. > > Here's my argument: the inequality describes a process by which we start > with 2 and take off half of what we had. Because we always take half, > we always leave half. Hence, we'll never get to zero, hence the > procedure never takes 2 units completely. > > I have also observed that > > 2  2^0 + 2^{1} + 2^{2} + ... + 2^{n} = 2^(n + 1) > > because if we always end up with half of what we had, that half will be > half of what we took last  after n halves, we took last 2^n and half > of that is just 1/2 2^n = 2^(n + 1). > > (*) On another exercise > > I asked a question in alt.algebra.help on April 6th 2016. I haven't > seen any activity there since then. I'll appreciate your attention on > the post ``on n^2  1 being divisible by 24'' whose MessageID is > <86oa9mtn6t.fsf@toledo.com>. If you find appropriate, please crosspost > your followup here so that others can consider it as well. Thank you!
Do you really need an induction proof? If you just sum the geometric series, you end up with
(12^(n+1))/(11/2) =(2^(n+1)1)/(2^(n+1)2^n =(2^(n+1)1)/2^n =22^(n) < 2.
If you really do need an induction proof, it should easily follow from the the last equality above.
Don



