Induction: $\sum_{k=1}^{2n} (-1)^k k = n$

Question

Use the proof of induction to show : $\sum_{k=1}^{2n} (-1)^k k = n$

I know how to show the base step of this problem, but in showing the inductive step I am having trouble determining how to show they are equal.

Answer 1

Suppose, it is true for $n-1$. Then $$\sum_{k=1}^{2n} (-1)^k k =$$ $$=\left( \sum_{k=1}^{2(n-1)} (-1)^k k\right)+(-1)^{2n-1}\cdot(2n-1)+(-1)^{2n}\cdot(2n)=$$ $$=(n-1)-(2n-1)+2n=n$$ it is also true for $n$.

Answer 2

let the relation holds for n<=m then in particular $\sum_{k=1}^{2m}(-1)^k k = m$ consider $\sum\limits_{k=1}^{2m+2}(-1)^k k = \sum\limits_{k=1}^{2m}(-1)^k k+ (-1)^{2m+1} (2m+1) + (-1)^{2m+2} (2m+2)=m-(2 m+1)+(2m+2)=m+1 $

Answer 3

$\begin{array}\\ \sum_{k=1}^{2n} (-1)^k k &=\sum_{k=1}^{n} ((-1)^{2k-1} (2k-1)+(-1)^{2k}(2k))\\ &=\sum_{k=1}^{n} (-(2k-1)+(2k))\\ &=\sum_{k=1}^{n} 1\\ \end{array} $

and you can either use induction to show that $\sum_{k=1}^{n} 1 = n $ (which might be a useful exercise) or you can just say that it is obvious.