Evaluating a summation $\sum_{k=0}^{100} k^2 + k$ I need help evaluating this:
$$\sum_{k=0}^{100} k^2 + k$$
I'm not sure how to solve this without writing the equation 100 times. and then adding them up.
 A: Since $(k+1)-k=1$ and $(k+1)^3-k^3=3k^2+3k+1$, we can write a telescoping sum:
$$
\begin{align}
\sum_{k=0}^n\left(k^2+k\right)
&=\sum_{k=0}^n\left[\frac13\left(3k^2+3k+1\right)-\frac13(1)\right]\\
&=\sum_{k=0}^n\left[\frac13\left((k+1)^3-k^3\right)-\frac13\left((k+1)-k\right)\right]\\
&=\frac13(n+1)^3-\frac13(n+1)
\end{align}
$$
A: Hint: What does $(k+1)^3 -k^3$ equal and what happens when you add this up as $k$ varies between one and one hundred?
A: It is known that:
$$\sum_{k=0}^{n} k=[(n^2 + n) / 2]$$
and
$$\sum_{k=0}^{n} k^2=[\frac{1}{6}n(n+1)(2n+1)]$$
Adding:
$$\sum_{k=0}^{n} k^2 + k$$
$$=[\frac{1}{6}n(n+1)(2n+1)] + [(n^2 + n) / 2]$$
In your case, $n=100$, so substitute to get:
$$((1/6)100(101)(201))+(1/2)(10000+100)=343400$$.
The solution provided by @robjohn is much more elegant of course!
If you are into sum history and some neat mathematical logic about sums, you may like this:Young Gauss and the sum of the first n positive integers.
A: Yet another approach is to apply the Euler-Mclaurin Sum Fromula.  
To that end, let $f(x)=x^2+x$.  Then, $f'(x)=2x+1$, $f''(x)=2$, and all higher-order derivatives are zero.  
Therefore, we can write
$$\begin{align}
\sum_{k=1}^nf(n)&=\int_0^nf(x)\,dx+\frac12(f(n)-f(0))+\frac{1}{12}(f'(n)-f'(0))\\\\
&=\frac{n^3}{3}+\frac{n^2}{2}+\frac12(n^2+n)+\frac16 n\\\\
&=\bbox[5px,border:2px solid #C0A000]{\frac13 n(n+1)(n+2)}
\end{align}$$
as expected!
A: $$\sum_{k=0}^{100} k^2+k=2\sum_{k=0}^{100}\binom {k+1}2=2\binom {102}3
=\frac{102\cdot 101\cdot 100}3\qquad\blacksquare$$
