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Prove by induction a property of the alternating sum of odd integers

Prove that

    \[ \sum_{k=1}^{2n} (-1)^k (2k+1) = 2n. \]

This implies the sum is proportional to n with constant of proportionality 2.


Proof. The proof is by induction. For the case n = 1 we have, on the left,

    \[ \sum_{k=1}^{2n} (-1)^k (2k+1) = \sum_{k=1}^2 (-1)^k (2k+1) = -3 + 5 = 2. \]

On the right we have 2n = 2. Hence, the formula holds for this case.
Assume then that the formula holds for some n = m \in \mathbb{Z}_{>0}. Then,

    \begin{alignat*}{2}  \sum_{k=1}^{2m} (-1)^k (2k+1) = 2m \qquad &\implies & \sum_{k=0}^{2(m+1)} (-1)^k (2k+1) &= 2m - (2(2m+1)+1) + (2(2m+2)+1) \\ &&&= 2m - 4m - 3 + 4m + 5 \\ &&&= 2(m+1). \end{alignat*}\end{align*}

Thus, if the statement is true for m then it is true for m+1. Hence, we have established the statement is true for all n \in \mathbb{Z}_{>0}. \qquad \blacksquare

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