I am trying to prove that $\sum_{i=1}^n i^4 = \frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} - \frac{n}{30}$.
I will use these formulas:
$\sum_{i=1}^n i = \frac{n(n+1)}{2} $
$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} $
$\sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}{4} $
I know I can use a system of equations for the proof, but I'm trying to come up with a different way.
I noticed that we can rewrite $\sum_{i=1}^m i^2$ as $\sum_{i=1}^n i^4$ (where $m = n^2$) plus some other terms.
For instance, $\sum_{i=1}^{16} i^2 = 1^2 + (2^2 + 3^2) + 4^2 + (5^2 + 6^2 + 7^2 + 8^2) + 9^2 + (10^2 + 11^2 + 12^2 + 13^2 + 14^2 + 15^2) + 16^2 = \sum_{i=1}^{4} (i^2)^2 + \sum_{i=1}^{4}[\sum_{j=1}^{2i-2}(j + (i-1)^2)^2]$.
For any n, $\sum_{i=1}^{n^2} i^2 = \sum_{i=1}^{n}[\sum_{j=1}^{2i-2}[(j + (i-1)^2)^2] + i^4]$.
Knowing that $(j + (i - 1)^2)^2 = j^2 + 2ji^2 - 4ji +2j + i^4 - 4i^3 + 6i^2 - 4i + 1$, we write:
$\sum_{i=1}^{n^2} i^2 = \sum_{i=1}^{n}[\sum_{j=1}^{2i-2} j^2 + \sum_{j=1}^{2i-2} (2i^2 - 4i + 2)j + (2i^4 - 4i^3 + 6i^2 - 4i + 1)$].
Evaluating the inner sums give us:
$\sum_{i=1}^{n^2} i^2 = \sum_{i=1}^{n}[\frac{(2i-2)(2i-1)(4i-3)}{6} + (2i^2 -4i + 2)\frac{(2i-2)(2i-1)}{2} + (2i^4 - 4i^3 + 6i^2 - 4i + 1)]$.
Simplifying the summand and isolating the $\sum_{i=1}^{n}i^4$ term gives us:
$6\sum_{i=1}^{n}i^4 = \sum_{i=1}^{n^2} i^2 - \sum_{i=1}^{n}(-\frac{46i^3}{3} + 18i^2 - \frac{29i}{3} + 2 )$
Carrying out the sums leads to:
$\sum_{i=1}^{n}i^4 = \frac{1}{6}[\frac{n^2(n^2+1)(2n^2+1)}{6} - (6n^3 + 9n^2 + 5n + \frac{-23n^4 - 46n^3 -52n^2 - 29n}{6})]$
The right side simplifies to $\frac{2n^6 + 26n^4 + 10n^3 - n^2 -n}{36}$.
I'm wondering if there is something wrong with the set-up given that our equation should have degree 5, but the $\sum_{i=1}^{n^2} i^2$ term ensures that we have degree 6. Does anyone have any thoughts? Thank you.
