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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 My algebra sucks. It turns into a giant mess!

These sorts of things seem to require a lot of intuition. (or, whats that word.... practice?)

Thank you very much!

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# Homework Help: Summation proof

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