- #1
Moridin
- 692
- 3
Homework Statement
Show that
\forall n \in \mathbb{N}:~ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} = \sum_{k=0}^{n} (-1)^{n+k} k^{2}
The Attempt at a Solution
I tried to apply the general method of proof by induction.
(1) Show that it is true for n = 1
\sum_{k=1}^{1} k = \frac{1(1+1)}{2} = \sum_{k=0}^{1} (-1)^{1+k} k^{2} = 1
(2) Show that if it is true for n = p, it is true for n = p + 1
Assume that
\sum_{k=1}^{n} k = \frac{n(n+1)}{2}
Now,
\sum_{k=1}^{p} k = 1 + 2 + 3 + ... + p
\sum_{k=1}^{p+1} k = 1 + 2 + 3+ ... + p + (p +1) = \sum_{k=1}^{p} k + (p+1)
According to our assumption, it follows that
\sum_{k=1}^{p} k + (p+1) = \frac{n(n+1)}{2} + (p + 1)
If it can be show that
\frac{p(p+1)}{2} + (p + 1) = \frac{(p+1)(p+2)}{2}
then the first equality is taken care of satisfactory. A trivial calculation shows that this is indeed the case.
For the second equality, the assumption is that
\sum_{k=0}^{p} (-1)^{p+k} k^{2} = \frac{p(p+1)}{2}
Here is it where it gets a bit tricky for me.
Now,
\sum_{k=0}^{p} (-1)^{p+k} k^{2} = (-1)^{p+1} + (-1)^{p+1} \cdot 4 + (-1)^{p+2} \cdot 8 + ... + (-1)^{2p} \cdot p^{2}
\sum_{k=0}^{p+1} (-1)^{p+k} k^{2} = (-1)^{p+1} + (-1)^{p+1} \cdot 4 + (-1)^{p+2} \cdot 8 + ... + (-1)^{2p} \cdot p^{2} + (-1)^{2p+2} \cdot (p+1)^{2} = \sum_{k=0}^{p} (-1)^{p+k} k^{2} + (-1)^{2p+2} \cdot (p+1)^{2}
Given our assumption, we get
\sum_{k=0}^{p} (-1)^{p+k} k^{2} + (-1)^{2p+2} \cdot (p+1)^{2} = \frac{p(p+1)}{2} + (-1)^{2p+2} \cdot (p+1)^{2}
So if it can be shown that
\frac{p(p+1)}{2} + (-1)^{2p+2} \cdot (p+1)^{2} = \frac{(p+1)(p+1)}{2}
then the second inequality is also taken care of satisfactory and part (2) is done.
\frac{p(p+1)}{2} + (-1)^{2p+2} \cdot (p+1)^{2} = \frac{p(p+1)}{2} + \frac{2(p+1)^{2}}{2} = \frac{p^{2} + p + 2(p^{2} +2p+1)}{2} = \frac{p^{2} + p + 2p^{2} + 4p + 2}{2}
But this does not equal \frac{(p+1)(p+1)}{2}, so I must have made one or more mistakes somewhere, but I am at a loss as to where it is.
