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Prove that 2^n can be expressed as sum of two consecutive numbers |
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Answer» Integers starting with n is kn+k(k−1)2 so if mcan be written in this form, then 2m=k(2n+k−1). If m is a power of 2, then both k and 2n+k−1 MUST be powers of two, but they have DIFFERENT PARITIES, so ONE of them must be 1. Presumably, k>1, so 2n+k−1=1 or 2n+k=2. Since k≥2, this MEANS that n≤0. |
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