# Math Help - Induction

1. ## Induction

I want to prove $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}=
\sum_{k=0}^{n}(-1)^{n+k}*k^2$

So, simply what I chose to do is to first prove
$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$
by (mathematical) induction, and then use the same method for
$\frac{n(n+1)}{2}=\sum_{k=0}^{n}(-1)^{n+k}*k^2$
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:
$\sum_{k=0}^{m_0+1}(-1)^{m_0+1+k}*k^2$
I then do my calculations and I get:

$\frac{3m_0+5m_0+2}{2}$
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).

2. Originally Posted by a4swe
I want to prove $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}=
\sum_{k=0}^{n}(-1)^{n+k}*k^2$

So, simply what I chose to do is to first prove
$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$
by (mathematical) induction, and then use the same method for
$\frac{n(n+1)}{2}=\sum_{k=0}^{n}(-1)^{n+k}*k^2$
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:
$\sum_{k=0}^{m_0+1}(-1)^{m_0+1+k}*k^2$
I then do my calculations and I get:

$\frac{3m_0+5m_0+2}{2}$
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 $m_0$ that:

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

Then:

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

$
=-\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 $-\frac{m_0(m_0+1)}{2}$ so:

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

RonL

3. Well then, there is something that I don not understand here:

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

and not:

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

4. Originally Posted by a4swe
Well then, there is something that I don not understand here:

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

and not:

$
=\sum_{k=0}^{m_0}(-1)^{m_0+1}*k^2 + (-1)^{2(m_0+1)}(m_0+1)^2
$
Because the first of these is the far right term in the original problem
with $n$ replaced by $m_0+1$, that is the next case after the one
assumed to hold.

RonL

5. Oh, sorry a little typo again I mean to say:

Why:

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

And not:

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

'cuse I did take the m_0+1 term for it self and only got m_0 left...right?

6. Originally Posted by a4swe
Oh, sorry a little typo again I mean to say:

Why:

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

And not:

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

'cuse I did take the m_0+1 term for it self and only got m_0 left...right?
It's the same answer as before. The first of these is:

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

with $m_0$ replaced by $m_0+1$ throughout, and
the last term taken outside the sum:

$
\sum_{k=0}^{m_0+1}(-1)^{m_0+1+k} k^2=\left[ \sum_{k=0}^{m_0}(-1)^{m_0+1+k} k^2 \right] +$
$(-1)^{m_0+1+m_0+1} (m_0+1)^2
$

RonL

7. Well, yes off course.
Hmmm, but why do one use n+k?
Isn't there any way to express the same thing with fix numbers, for me it seems like that should be possible, but probably I am wrong.
Thanks a lot CB, you are often of great help.