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?


The sum of the $k$-th powers of    numbers   $1$ to $n$ can also be derived by means of formal power series. 
  \begin{align*}
S_k(n):=\sum_{j=1}^n j^k\qquad\qquad n,k\geq 1
\end{align*}
  In order to do so we encode for fixed $k$ the sequence
  $\left(S_k(n)\right)_{n\geq 0}$ by generating functions $$A(z)=\sum_{n=0}^\infty S_k(n)z^n$$

It's convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. We also use $D_z:=\frac{d}{dz}$ to denote  the differential operator.   The following is valid:  

The sum of the $k$-th powers of numbers $1$ to $n$ is given by
  \begin{align*}
S_k(n)=\sum_{j=1}^nj^k=[z^n]\frac{1}{1-z}(zD_z)^k\frac{1}{1-z}\tag{1}
\end{align*}
A  detailed  derivation of (1) together     with some examples  can     be found in this answer.

Here is an example with $k=4$

Example $S_4(n)$
\begin{align*}
S_4(n)=\sum_{j=0}^n j^4&=[z^n]\frac{1}{1-z}(zD_z)^4\frac{1}{1-z}\tag{2}\\
&=[z^n]\frac{z(1+11z+11z^2+z^3)}{(1-z)^6}\\
&=[z^n](z+11z^2+11z^3+z^4)\sum_{j=0}^{\infty}\binom{-6}{j}(-z)^{j}\tag{3}\\
&=\left([z^{n-1}]+11[z^{n-2}]+11[z^{n-3}]+[z^{n-4}]\right)\sum_{j=0}^{\infty}\binom{j+5}{5}z^j\tag{4}\\
&=\binom{n+4}{5}+11\binom{n+3}{5}+11\binom{n+2}{5}+\binom{n+1}{5}\\
&=\frac{1}{30}n(6n^4+15n^3+10n^2-1)
\end{align*}

Comment:


*

*In (2) we apply the operator $\frac{1}{1-z}(zD_z)^4$ to $\frac{1}{1-z}$

*In (3) we use the binomial series expansion

*In (4) we use         the linearity    of the coefficient of operator, apply the formula
\begin{align*}
  [z^{p-q}]A(z)=[z^p]z^qA(z)
  \end{align*}
and use the binomial identity
\begin{align*}
  \binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q
  \end{align*}

*In (5) we select the coefficients of $z^{n-1}$   to $z^{n-4}$.

