How to show this statement by induction

[Moridin]

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.

 

[GregA]

when you split the sum $$\sum_{k=0}^{p+1}(-1)^{k+1+p}k^{2}$$ you should find you can get [tex]-(\sum_{k=0}^{p}k) + (-1)^{2p +2}(p+1)^2

 

[tiny-tim]

Hi Moridin!

Alternatve proof for second part …

Hint: what is (2k)² - (2k - 1)² ?

 

[Moridin]

Hi Moridin!

Alternatve proof for second part …

Hint: what is (2k)² - (2k - 1)² ?

4k - 1?

You lost me.

 

[GregA]

Sorry about my last advice...I didn't identify exactly where you had gone wrong.

Your mistake stems from this line:
$$\sum_{k=0}^{p+1} (-1)^{p+k} k^{2} = (-1)^{p+1} + (-1)^{p+1} \cdot 4 + (-1)^{p+2} \cdot 8 + ... + (-1)^{2p} \cdot p^{2} + (-1)^{2p+2} \cdot (p+1)^{2} = \sum_{k=0}^{p} (-1)^{p+k} k^{2} + (-1)^{2p+2} \cdot (p+1)^{2}$$

you should be starting with $$\sum_{k=0}^{p+1} (-1)^{p+1+k} k^{2} =...$$ not $$\sum_{k=0}^{p+1} (-1)^{p+k} k^{2} = ...$$

 

[Defennder]

Can you show that $$\sum^{n+1}_{k=0} (-1)^{n+k+1}k^2 = -\sum^{n}_{k=0} (-1)^{n+k} k^2 \ + \ (-1)^{2(n+1)} (n+1)^2$$?

Note that the last term of the summation is always positive, and given that the terms alternate between positive and negative, this amounts to multiplying the entire summation by -1 to "shift" the summation backwards by one index to accomodate the last summand which must be positive.

 

[tiny-tim]

4k - 1?

You lost me.

Hint: split 4k - 1 into two parts.

 

[Moridin]

Can you show that $$\sum^{n+1}_{k=0} (-1)^{n+k+1}k^2 = -\sum^{n}_{k=0} (-1)^{n+k} k^2 \ + \ (-1)^{2(n+1)} (n+1)^2$$?

Note that the last term of the summation is always positive, and given that the terms alternate between positive and negative, this amounts to multiplying the entire summation by -1 to "shift" the summation backwards by one index to accomodate the last summand which must be positive.

I understand that equality now, since the multiplier of negative one shifts the exponents one step back according the the power laws. Here is what I got so far.

$$\sum^{p+1}_{k=0} (-1)^{p+k+1}k^2 = -\sum^{p}_{k=0} (-1)^{p+k} k^2 \ + \ (-1)^{2(p+1)} (p+1)^2 = - \frac{p(p+1)}{2} \ + \ (-1)^{2(p+1)} (p+1)^2$$

However,

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

I seem to have made a sign error, because if the last three subtractions where additions, I would be home free.

 

[tiny-tim]

Hint: split 4k - 1 into two parts.

Hint: 4k - 1 = 2k + (2k - 1).

So ∑(4k - 1) over even k = … ?

 

[Moridin]

Hint: 4k - 1 = 2k + (2k - 1).

So ∑(4k - 1) over even k = … ?

7 + 15 + 23 + ... + ?

 

[tiny-tim]

7 + 15 + 23 + ... + ?

Well, yes, but we're trying to avoid summing that, since it's too difficult.

It equals ∑2k + ∑(2k - 1) over even k, which we can express as a sum over all k, as … ?

 

[GregA]

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

I seem to have made a sign error, because if the last three subtractions where additions, I would be home free.

$$- \frac{p(p+1)}{2} \ + \ (-1)^{2(p+1)} (p+1)^2 \neq - \frac{p(p+1)}{2} \ - \ (p+1)^2$$

 

[Moridin]

$$- \frac{p(p+1)}{2} \ + \ (-1)^{2(p+1)} (p+1)^2 \neq - \frac{p(p+1)}{2} \ - \ (p+1)^2$$

I thought I made a mistake around there. Oh right, it is 2m + 1 that gives an odd exponent, but 2m+ 2 gives an even exponent for all positive m? So it should actually be

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

That simplifies to what we're looking for namely

$$\frac {(p+1)(p+2)}{2}$$

Then it just a matter of applying the axiom of induction right?

 

[GregA]

I thought I made a mistake around there. Oh right, it is 2m + 1 that gives an odd exponent, but 2m+ 2 gives an even exponent for all positive m? So it should actually be

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

That simplifies to what we're looking for namely

$$\frac {(p+1)(p+2)}{2}$$

Then it just a matter of applying the axiom of induction right?

indeed

 

[Moridin]

Thank you all so much for your time and help!