Proving a Sum of Powers of Integers: A Challenge!

[gazzo]

Hey!

Can someone please give me a hint on this

Prove:
$$\sum_{n=1}^n i^4 = \frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30}$$

What I've got so far:

Let $P(n)$ be the statement:
$$\sum_{n=1}^n i^4 = \frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30}$$

Let $n=1$ we get;
$$\sum_{n=1}^1 i^4 = \frac{1(1+1)(2(1)+1)(3(1^2) + 31 - 1)}{30}<br /> = \frac{(2)(3)(5)}{30}<br /> = 1$$
Which is true.

Assume $P(n)$ is true $\forall k \ge n, k \in\mathbb{Z}$
Let $n=k+1$

Then we get:
$$P(k+1) = \sum_{n=1}^{k+1} i^4 = \bigg( \sum_{n=1}^{k} i^4 \bigg) + (k+1)^4 = \frac{k(k+1)(2k+1)(3k^2 + 3k - 1)}{30} + (k+1)^4<br /> = \frac{1}{30} \bigg[ k(k+1)(2k+1)(3k^2+3k-1) + 30(k+1)^4 \bigg]$$

and then i tried
$$\frac{1}{30} \bigg[ k(k+1)(2k+1)(3k^2+3k-1) + 30(k+1)^4 \bigg] = \frac{k+1}{30} \bigg[ k(2k+1)(3k^2+3k-1) + 30(k+1)^3 \bigg]$$

A well as heaps of other arrangements My algebra sucks. It turns into a giant mess!
These sorts of things seem to require a lot of intuition. (or, what's that word... practice?)

Thank you very much!

 

[HallsofIvy]

Good! Factoring out (k+1) simplifies it. I'm afraid what you are going to have to do now is actually multiply that out:
multiply (2k)(2k+1)(3k²+ 3k- 1) multiply 30(k+1)³ and add them
Now factor that. It may help to know that you WANT (since you have already factored out (k+1) (which corresponds to the "n" in the original formula) (k+2)(2k+3)(3k²+ 9k+ 5) which correspond, respectively, to n+1, 2n+1, and 3n²+ 3n-1 with k+1 in place of n.

 

[gazzo]

Thanks HallsofIvy

Hmm...

$$\frac{1}{30} \bigg[ k(k+1)(2k+1)(3k^2+3k-1) + 30(k+1)^4 \bigg] = \frac{k+1}{30} \bigg[ k(2k+1)(3k^2+3k-1) + 30(k+1)^3 \bigg]$$

$$= \frac{k+1}{30} \bigg[ k(6k^3 + 6k^2 - 2k + 3k^2 + 3k - 1) + 30(k+1)^3 \bigg]$$

$$= \frac{k+1}{30} \bigg[ k(6k^3+9k^2+k-1) + 30(k+1)^3 \bigg]$$

$$= \frac{k+1}{30} \bigg[ (6k^4+9k^3+k^2-k) + 30(k+1)^3 \bigg]$$

$$= \frac{k+1}{30} \bigg[ (6k^4+9k^3+k^2-k) + 30(k+1)(k^2+2k+1) \bigg]$$

$$= \frac{k+1}{30} \bigg[ (6k^4+9k^3+k^2-k) + 30(k^3+2k^2+k+k^2+2k+1) \bigg]$$

$$= \frac{k+1}{30} \bigg[ (6k^4+9k^3+k^2-k) + (30k^3+60k^2+30k+30k^2+60k+30) \bigg]$$

$$= \frac{k+1}{30} \bigg[ (6k^4+9k^3+k^2-k) + (30k^3+90k^2+90k+30) \bigg]$$

$$= \frac{k+1}{30} \bigg[ (6k^4+9k^3+k^2-k) + (30k^3+90k^2+90k+30) \bigg]$$

$$= \frac{k+1}{30} \bigg[ 6k^4+39k^3+90k^2+90k+30 \bigg]$$

$$= \frac{k+1}{30} \bigg[ 3k(2k^3+13k^2) + 30(3k^2+3k+1) \bigg]$$

$$= \frac{k+1}{30} \bigg[ 3k(2k^3+13k^2) + 30(3k(k+1)+1) \bigg]$$

ack!

i tried getting it to look something like err at least..

$$\frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30} = \frac{(2n^2+3n+1)(3n^2+3n-1)}{30}$$

$$= \frac{6n^4+3n^3-2n^2+9n^3+9n^2-3n+3n^2+3n-1}{30}$$

$$= \frac{6n^4+12n^3+10n^2-1}{30}$$

Which, where $n=k$ :

$$\frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30} = \frac{(k+1)(k+2)(2k+3)(3k^2+9k+5)}{30}$$

:shy:

 

[Hurkyl]

It might help to multiply it out before doing anything. In other words, prove the statement in its non-factored form.

There's also a cheap trick you can use. If, for some reason, you know that the answer is a polynomial of degree 5, then you can just try 6 values of n and check you get the right answer. (given (k+1) input-output pairs, there is exactly one degree k polynomial passing through them)

 

[learningphysics]

$$= \frac{k+1}{30} \bigg[ 6k^4+39k^3+90k^2+90k+30 \bigg]$$

There's a mistake above... it should be 91k^2 and 89k.

I think Hurkyl's method of multiplying out what you're trying to prove...(k+2)(2k+3) etc... and showing that it's the same as the above polynomial is the best way.

If you want to factor your above polynomial ... then since you "expect" (k+2), and (2k+3) as factors... plug in -2 into the polynomial above to see that it goes to zero. That proves that k+2 is a factor. Also try pluggin in k=-3/2 proves that 2k+3 is a factor ince 2k+3= 2(k+3/2).

You can do polynomial division then to get the last factor.