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##### Prove by induction please

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Apr 24th, 2015

First: check the statement for the n=1. The left side is equal to (-1)^0*1^2 = 1, the right side (-1)^0*(1*2/2) is also equal to 1.

The induction step: suppose we know that the statement is true for some k>=1, prove that it is true for k+1.

Sum by i from 1 to (k+1) of [(-1)^(i-1)*i^2] = the same sum to i=k + (k+1)'s summand = (by the induction assumption)
= (-1)^(k-1)*[k(k+1)/2] + (-1)^k*(k+1)^2 = (-1)^(k-1)*[k(k+1)/2] + (-1)^(k-1)*(-1)*(k+1)^2.

This is equal to (-1)^(k-1)*(k+1)*[k/2 - (k+1)] =  (-1)^(k-1)*(k+1)*[-k/2 - 1] =
= (-1)^(k-1)*(k+1)*[(-1/2)*(k + 2)] = (-1)^k*[(k+1)(k+2)/2].

It is exactly the same we see at the right side for n=k+1, so the statement is proved for all n>=1.

Apr 24th, 2015

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Apr 24th, 2015
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Apr 24th, 2015
Aug 23rd, 2017
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