<s>[INST] Summation Theorem how to get formula for exponent greater than 3 I'm studying in the summer for calculus 2 in the fall and I'm reading about summation. I'm given these formulas:
\begin{align*} 
\sum_{i=1}^n 1 &= n, \\
\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 &=\left[ \frac{n(n+1)}{2}\right]^2.
\end{align*}
But how do I come up with the right formula for exponents greater than 3?
 [/INST] For any polynomial of $p$th powers  $S_p(n) = \sum_{k=1}^n k^p$ $(p \in \Bbb N)$ there is a formula of higher degree $S_p(n)=\sum_{k=1}^{p+1} a_kn^k$
The question is how to find the unknown $a_k$ coefficients which are real numbers. 
So use the following formulas to find them:
$a_k= \frac{1}{k!} \cdot \sum_{j=1}^{p+1-k} (-1)^{j+1} \frac {a_{k+j} \cdot (k+j)!}{(j+1)!}$ for $(0 \lt k \lt p+1)$ 
$a_k=\frac {1}{p+1}$ for $(k=p+1)$
Example:
$S_3(n) = \sum_{k=1}^n k^3$
Solution:
$p=3$
$S_3(n)=\sum_{k=1}^{4} a_kn^3 = a_1n+a_2n^2+a_3n^3+a_4n^4$
Now using the above two formulas to find the unknown coefficients $a_1,a_2,a_3,a_4$
$a_4 = \frac{1}{3+1}=\frac{1}{4}$
$a_3=\frac {1}{3!} \cdot \sum_{j=1}^{1} (-1)^{j+1} \frac {a_{3+j} \cdot (3+j)!}{(j+1)!} =\frac {1}{3!} \cdot \frac {a_4 \cdot 4!}{2!} = \frac {\frac{1}{4} \cdot 4!}{3! \cdot 2!}=\frac {1}{2}$
$a_2=\frac {1}{2!} \cdot \sum_{j=1}^{2} (-1)^{j+1} \frac {a_{2+j} \cdot (2+j)!}{(j+1)!} = \frac {1}{2!} \cdot (\frac {a_3 \cdot 3!}{2!} - \frac {a_4 \cdot 4!}{3!}) = \frac {1}{2} \cdot (3a_3-4a_4)= \frac {1}{2} \cdot (3 \cdot \frac {1}{2}-4 \cdot \frac {1}{4})= \frac {1}{4}$
$a_1=\frac {1}{1!} \cdot \sum_{j=1}^{3} (-1)^{j+1} \frac {a_{1+j} \cdot (1+j)!}{(j+1)!} =  \sum_{j=1}^{3} (-1)^{j+1} a_{1+j}=a_2-a_3+a_4 =\frac {1}{4}-\frac {1}{2}+\frac {1}{4}=0$
Finally
$S_3(n)=0*n+\frac {1}{4}*n^2+\frac {1}{2}*n^3+\frac {1}{4}*n^4 =\frac {n^2+2n^3+n^4}{4}=\left[ \frac{n(n+1)}{2}\right]^2.$
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