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Establish a formula for alternating sum of squares

by RoRi

Claim:

    \[ 1 - 4 + 9 - 16 + \cdots + (-1)^{n+1} n^2 = (-1)^{n+1}(1+ \cdots +n) \]

Proof. For n =1, we have 1 on the left, and on the right (-1)^2 (1) = 1. Thus, the formula is true for n =1.
Assume then that it is true for some n =k \in \mathbb{Z}_{>0}. Then,

    \begin{align*} 1-4+9- \cdots + (-1)^{k+1}k^2 &= (-1)^{k+1} (1+ \cdots +k) \\ \implies 1 + \cdots + (-1)^{k+2}(k+1)^2 &= (-1)^{k+1} (1+ \cdots + k) + (-1)^{k+2}(k+1)^2 &(\text{Adding } (-1)^{k+2}(k+1)^2) \\ &= (-1)^{k+2} \left(\frac{-k(k+1)}{2} \right) + (-1)^{k+2} (k+1)^2 \\ &= (-1)^{k+2} \left( k^2 + 2k +1 - \frac{k^2 + k}{2} \right) \\ &= (-1)^{k+2} \left( \frac{k^2}{2} + \frac{3k}{2} + 1 \right) \\ &= (-1)^{k+2} \left( \frac{(k+1)(k+2)}{2} \right) \\ &= (-1)^{k+2} (1+ \cdots + (k+1)) & (\text{I.4.4, Exercise #1 (a)}). \end{align*}

Thus, if the formula is true for k then it is true for k+1. Since we established it is true for n=1, we have that it is true for all n \in \mathbb{Z}_{>0}. \qquad \blacksquare

4 comments

    • Tiago says:

      Hi Luis. He used the identity: (1+2+…+k) = k(k+1)/2 (this appears in part (a) of the exercise); and he put the factor (-1)^(k+2) out. Notice that you have [(-1)^(k+1)] * [k(k+1)/2], so when you put (-1)^(k+2) out, what stays inside the parenthesis is: [(-1)^(-1)] * [k(k+1)/2]. But (-1)^(-1) = -1. Hence you get, – k(k+1)/2.

      Remark: remember that (a^b) * (a^c) = a^(b+c).

      • Anonymous says:

        Hello. In the 3rd step: How did you go from (-1)^(k+1) to (-1)^(k+2)?
        We have [(-1)^(k+1)] * [k(k+1)/2], but i do not see from where i can get (-1)^(k+2) and put out, and therefore have [(-1)^(-1)] * [k(k+1)/2]. Could you please explain?

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