Let a0,a1,…,an be integers, one of which is nonzero, and all of the numbers are not less than −1. Prove that if a0+2a1+22a2+⋯+2nan=0, then a0+a1+⋯+an>0.
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Tags: induction, algebra proposed, algebra
04.03.2009 22:54
Induction: 1. a0+2a1=0 ⇔(a0,a1)=(2,−1). (If a1≥0, then a0 is at least −1 but even−1 can't be 0.) 2. a0+2a1+22a2+⋯+2nan=0 ⇔a0+2(a1+2a2+⋯+2n−1an)=0 Thus a0 is even, so a0=2k for some integer k≥0 and therefore (a1+k)+2a2+⋯+2n−1an=0. Obviously, a1+k≥−1 and then by the induction hypothesis we have k+a1+a2+⋯+an>0, but 2k≥k ⇒a0+a1+⋯+an>0, too
08.10.2017 15:42
Also induction. 2|a0, so a0≥0,a0/2≥0. And we have (a0/2+a1)+2a2+…+2n−1an=0, a0/2+a1≥−1. So by induction hypothesis we get that a0/2+a1+a2+…+an≥0 and so (a0≥0) a0+…+an≥0. Done
08.06.2023 02:52
BG Yoda wrote: Induction: 1. a0+2a1=0 ⇔(a0,a1)=(2,−1). (If a1≥0, then a0 is at least −1 but even−1 can't be 0.) 2. a0+2a1+22a2+⋯+2nan=0 ⇔a0+2(a1+2a2+⋯+2n−1an)=0 Thus a0 is even, so a0=2k for some integer k≥0 and therefore (a1+k)+2a2+⋯+2n−1an=0. Obviously, a1+k≥−1 and then by the induction hypothesis we have k+a1+a2+⋯+an>0, but 2k≥k ⇒a0+a1+⋯+an>0, too the same solution i had