Home » Blog » Prove by induction a property of the alternating sum of odd integers

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

by RoRi

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

2 comments

    • Astute Chic says:

      He doesn’t start with the index k=0, it’s just a typo, a mistake.
      Then the induction step is where he has to evaluate the sum from k=1 to k=2(m+1), i.e. from k=1 to k=2m +2. We have already assumed the result of the sum from k=1 to k=2m, so he added the last to terms which evaluate the ‘+2’ part. So these two terms are the “n+1” terms you were talking about. There are two because 2(m+1)=2m +(2).

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):