Originally Posted by

**a4swe** I want to prove $\displaystyle \sum_{k=1}^{n}k=\frac{n(n+1)}{2}=

\sum_{k=0}^{n}(-1)^{n+k}*k^2$

So, simply what I chose to do is to first prove

$\displaystyle \sum_{k=1}^{n}k=\frac{n(n+1)}{2}$

by (mathematical) induction, and then use the same method for

$\displaystyle \frac{n(n+1)}{2}=\sum_{k=0}^{n}(-1)^{n+k}*k^2$

The first works out fine, no problem there at all.

The second part causes problems.

I manage to show P(x) true for x=1 (wow, I am realy amazing)

And then I make my claim:

$\displaystyle \sum_{k=0}^{m_0+1}(-1)^{m_0+1+k}*k^2$

I then do my calculations and I get:

$\displaystyle \frac{3m_0+5m_0+2}{2}$

Witch is not right.

Anyone know or have a clue about where I am off?

Is there another part of my calculations that I should include? Please say so if that is the case, I try to include just the parts where I think I am wrong (everything only if I have no clue).

Try:

Assume for some integer $\displaystyle m_0$ that:

$\displaystyle

\frac{m_0(m_0+1)}{2}=\sum_{k=0}^{m_0}(-1)^{m_0+k}*k^2

$

Then:

$\displaystyle

\sum_{k=0}^{m_0+1}(-1)^{m_0+1+k}*k^2$$\displaystyle

=\sum_{k=0}^{m_0}(-1)^{m_0+1+k}*k^2 + (-1)^{2(m_0+1)}(m_0+1)^2

$

$\displaystyle

=-\sum_{k=0}^{m_0}(-1)^{m_0+k}*k^2+(m_0+1)^2

$

.

Then by assumption the first term on the right may be replaced by $\displaystyle -\frac{m_0(m_0+1)}{2}$ so:

$\displaystyle

\sum_{k=0}^{m_0+1}(-1)^{m_0+1+k}*k^2$$\displaystyle

=-\frac{m_0(m_0+1)}{2}+(m_0+1)^2

$.

Which should simplify to what you are looking for to prove the required equality for $\displaystyle m_0+1$.

RonL