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Base Case:
\begin{eqnarray*} \sum_{k=1}^{2n} (-1)^k k = n\\ (-1)^1 (1) + (-1)^2(2) &=&1 \\ 1=1 \end{eqnarray*}

Inductive Step: For this step we must prove that \begin{eqnarray*} \sum_{k=1}^{2n} (-1)^k k = n \Rightarrow \sum_{k=1}^{2(n+1)} (-1)^k k = n+1 \\ \sum_{k=1}^{2(n+1)} (-1)^k k = \sum_{k=1}^{2n} (-1)^k k + \sum_{k=1}^{2} (-1)^k k \end{eqnarray*} We know that, \begin{eqnarray*} \sum_{k=1}^{2n} (-1)^k k = n \end{eqnarray*} and from the base case we conclude, \begin{eqnarray*} \sum_{k=1}^{2} (-1)^k k = 1 \end{eqnarray*} Therefore in we have $n+1 = n+1$. Is this a method that could be utilized?

user26857
  • 52,094
Katie Loss
  • 59

2 Answers2

3

Basically, yes. There is a bit of a snag with your second summation though: it should be $k = 2n + 1$ and $k = 2n + 2$. So we have: \begin{align*} \sum_{k=1}^{2(n + 1)} (-1)^k k &= \sum_{k=1}^{2n} (-1)^k k + (-1)^{2n + 1}(2n + 1) + (-1)^{2n + 2}(2n + 2) \\ &= n + (-1)^{2n + 1}(2n + 1) + (-1)^{2n + 2}(2n + 2) &\text{by the induction hypothesis}\\ &= n - (2n + 1) + (2n + 2)\\ &= n + 1\\ \end{align*} as desired.

Adriano
  • 41,576
1

For the record, the identity can be proved without induction by combining pairs of consecutive terms:

$$\begin{align*} \sum_{k=1}^{2n}(-1)^kk&=\sum_{k=1}^n\Big((-1)^{2k-1}(2k-1)+(-1)^{2k}(2k)\Big)\\ &=\sum_{k=1}^n\Big(-(2k-1)+2k\Big)\\ &=\sum_{k=1}^n1\\ &=n\;. \end{align*}$$

Brian M. Scott
  • 616,228