Proof by induction $\sum_{k=1}^{2n}(-1)^{k-1}\cdot k=-n$

Question

Hi I am trying to proof that $\sum_{k=1}^{2n}(-1)^{k-1}\cdot k=-n$ by induction.

I end up here: $-n+(-1)^{2n+2-1} \cdot (2n+2) = -n -2n -2 \ne -n$

Could someone help me find my mistake?

Thanks!

Answer 1

For $n=1$, \begin{equation*} \sum_{k=1}^{2}(-1)^{k-1}\cdot k=(-1)^1\cdot1=-1. \end{equation*} Suppose it holds for $n$, i. e., \begin{equation}\label{1} \sum_{k=1}^{2n}(-1)^{k-1}\cdot k=-n. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) \end{equation} Let's prove the result for $n+1$, \begin{equation} \begin{aligned} \sum_{k=1}^{2(n+1)}(-1)^{k-1}\cdot k & = \sum_{k=1}^{2n+2}(-1)^{k-1}\cdot k \ \ \ \ \ \text{(we separate the last two terms of the sum)}\\ & =\sum_{k=1}^{2n}(-1)^{k-1}\cdot k+(-1)^{(2n+1)-1}\cdot(2n+1)+(-1)^{(2n+2)-1}\cdot(2n+2)\\ & =\underbrace{-n}_{\text{For} (1)}+(-1)^{2n}\cdot(2n+1)+(-1)^{2n+1}\cdot(2n+2)\\ & =-n+2n+1+(-1)(2n+2)\\ &=-n+2n+1-2n-2=-n-1\\ &=-(n+1). \end{aligned} \end{equation}

Answer 2

Plug in $n=1$ and see that the sum is $ -1 $. Now, assume the property holds for some $n$. From this one is to show that it implies that the property holds for $n+1$. Notice that if $n=1$, the formula must hold for $n=2$. But since it holds for $n=2$, it must also be true for $n=3$, and so on. Hence the name: induction.

Proof that if $\sum_{k=1}^{2n} (-1)^{k-1}k = -n$ then $\sum_{k=1}^{2(n+1)} (-1)^{k-1}k$:

$$ \sum_{k=1}^{2(n+1)} (-1)^{k-1}k = \sum_{k=1}^{2n} (-1)^{k-1}k + (2n+1) - (2n-2) = -n + 2n + 1 - 2n - 2 = -n - 1 = -(n+1). $$ This is because of the induction hypothesis. Therefore the property holds for all positive integers $n$.