Induction

**a4swe** · September 4th 2006, 12:47 PM

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).

**CaptainBlack** · September 4th 2006, 01:31 PM

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

**a4swe** · September 4th 2006, 10:15 PM

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 $

**CaptainBlack** · September 4th 2006, 10:32 PM

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

**a4swe** · September 4th 2006, 10:42 PM

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?

**CaptainBlack** · September 4th 2006, 10:58 PM

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

**a4swe** · September 4th 2006, 11:26 PM

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.

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