<s>[INST] Easy way of memorizing or quickly deriving summation formulas My math professor recently told us that she wants us to be familiar with summation notation. She says we have to have it mastered because we are starting integration next week. She gave us a bunch of formulas to memorize. I know I can simply memorize the list, but I am wondering if there is a quick intuitive way of deriving them on the fly. It has to be a really quick derivation because all of her test are timed.
Otherwise is there an easy way, you guys remember these formulas. 
$\begin{align}
\displaystyle
&\sum_{k=1}^n k=\frac{n(n+1)}{2} \\
&\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6} \\
&\sum_{k=1}^n k^3=\frac{n^2(n+1)^2}{4} \\
&\sum_{k=1}^n k(k+1)=\frac{n(n+1)(n+2)}{3} \\
&\sum_{k=1}^n \frac{1}{k(k+1)}=\frac{n}{n+1} \\
&\sum_{k=1}^n k(k+1)(k+2)=\frac{n(n+1)(n+2)(n+3)}{4} \\
&\sum_{k=1}^n \frac{1}{k(k+1)(k+2)}=\frac{n(n+3)}{4(n+1)(n+2)} \\
&\sum_{k=1}^n (2k-1)=n^2
\end{align}$
Note: Sorry if there is an easy and obvious answer to this question. Most of the students in my class already know these formulas from high school, but I only went up to Algebra 2 and Trig when I was in high school. PS: This is for a calculus class in college.
 [/INST] Perhaps reorganizing them this way :
Class 1 : clear pattern (without end in fact!)
\begin{align}
\sum_{k=1}^n k&=\frac{n(n+1)}2\\
\sum_{k=1}^n k(k+1)&=\frac{n(n+1)(n+2)}3\\
\sum_{k=1}^n k(k+1)(k+2)&=\frac{n(n+1)(n+2)(n+3)}4\\
&\cdots\\
\end{align}
$$\boxed{\displaystyle\sum_{k=1}^n k^{(p)}=\frac{n^{(p+1)}}{p+1}}$$
(the rising factorial $\;k^{(p)}:=k(k+1)(k+2)\cdots(k+p-1)\;$ and the analogy between sum and integration should help to remember all these and... make nice mathematics!)
Class 2 : using simple relations as $\,\sum k^2=\sum k(k+1)-\sum k\;$ and previous results...
\begin{align}
\sum_{k=1}^n k&=\frac{n(n+1)}2\\
\sum_{k=1}^n k^2&=\frac{n(n+\frac 12)(n+\frac 22)}3\\
\\
\sum_{k=1}^n k^3&=\left(\sum k\right)^2\\
\end{align}
Class 3 : reciprocal rising factorial :
\begin{align}
\sum_{k=1}^n \frac 1{k(k+1)}&=\frac 1{1}-\frac 1{n+1}\qquad \text{}\\
\sum_{k=1}^n \frac 2{k(k+1)(k+2)}&=\frac 1{1\cdot 2}-\frac 1{(n+1)(n+2)}\\
&\cdots\\
\end{align}
$$\boxed{\displaystyle\sum_{k=1}^n \frac p{k^{(p+1)}}=\frac 1{1^{(p)}}-\frac 1{(n+1)^{(p)}}}\qquad 1^{(p)}=p!$$
Now remember that $\quad \displaystyle\int_{k=1}^{n+1} \frac p{k^{p+1}}\,dk=\frac 1{1^p}-\frac 1{(n+1)^p}$
(our upper-bound had to be changed from $n\to n+1$ while for class $1$ the lower-bound $1\to 0$)
Concerning the last one (sum of odds) remember this picture :
 $\qquad\displaystyle\sum_{k=1}^n (2k-1) =n^2\quad $
</s>