Proving a Sum of Powers of Integers: A Challenge!

In summary: I get (3k^2 + 3k - 1). So... if you get that (k+2)(2k+3)(3k^2 + 3k - 1) = 6k^4+12k^3+10k^2-k. Then you've proven it. Hope that helps.
  • #1
gazzo
175
0
Hey!

Can someone please give me a hint on this :uhh:

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

What I've got so far:

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

Let [itex]n=1[/itex] we get;
[tex]
\sum_{n=1}^1 i^4 = \frac{1(1+1)(2(1)+1)(3(1^2) + 31 - 1)}{30}
= \frac{(2)(3)(5)}{30}
= 1
[/tex]
Which is true.

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

Then we get:
[tex]
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
= \frac{1}{30} \bigg[ k(k+1)(2k+1)(3k^2+3k-1) + 30(k+1)^4 \bigg]
[/tex]

and then i tried
[tex]
\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]
[/tex]

A well as heaps of other arrangements :frown: :cry: 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! :redface: :redface: :blushing:
 
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  • #2
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.
 
  • #3
Thanks HallsofIvy :smile:

Hmm...

[tex]\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][/tex]

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

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

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

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

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

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

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

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

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

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

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

ack!

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

[tex]\frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30} = \frac{(2n^2+3n+1)(3n^2+3n-1)}{30}[/tex]

[tex]= \frac{6n^4+3n^3-2n^2+9n^3+9n^2-3n+3n^2+3n-1}{30}[/tex]

[tex]= \frac{6n^4+12n^3+10n^2-1}{30}[/tex]

Which, where [itex] n=k [/itex] :

[tex]\frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30} = \frac{(k+1)(k+2)(2k+3)(3k^2+9k+5)}{30}[/tex]

:shy:
 
  • #4
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. :biggrin: 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)
 
  • #5
gazzo said:
[tex]= \frac{k+1}{30} \bigg[ 6k^4+39k^3+90k^2+90k+30 \bigg][/tex]

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.
 

1. What is the definition of "Proving a Sum of Powers of Integers"?

Proving a Sum of Powers of Integers is a mathematical challenge that involves finding a specific set of integers that, when added together, equal a given number raised to a certain power. For example, proving that 1^2 + 2^2 + 3^2 = 14.

2. How difficult is it to solve the challenge of "Proving a Sum of Powers of Integers"?

The difficulty of this challenge varies depending on the specific numbers and powers involved. Some cases may be relatively simple to solve, while others may require advanced mathematical techniques and multiple steps to prove.

3. Are there any specific strategies or techniques that can be used to solve this challenge?

Yes, there are various strategies and techniques that can be used to solve the challenge of proving a sum of powers of integers. These include algebraic manipulation, factoring, and identifying patterns in the numbers.

4. Is there a limit to the number of integers that can be used in the sum?

No, there is no limit to the number of integers that can be used in the sum. However, the more integers involved, the more difficult the challenge may become.

5. How is "Proving a Sum of Powers of Integers" relevant in real-world applications?

While this challenge may not have direct real-world applications, it can help develop critical thinking and problem-solving skills that are useful in various fields, including mathematics, physics, and computer science.

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