Question

How to find the \[ \sum_{i = 0}^{n}i^2 \]...
that is the sum of i^2 for values i = 0 -> n? Is there a trick for this or must each term be summed individually?

Answer 1

Use induction

Answer 2

\[
\sum _{i=0}^n i^2=\frac{1}{6} n (n+1) (2 n+1)
\]

Answer 3

It is true for n=0. Suppose it is true for n and prove it for n+1

Answer 4

\[
\sum _{i=0}^{n+1} i^2=\frac{1}{6} n (n+1) (2 n+1)+(n+1)^2 =\\
(n+1) \left(\frac{1}{6} n (2 n+1)+(n+1) \right)
\]

Answer 5

\[
\sum _{i=0}^{n+1} i^2=\frac{1}{6} n (n+1) (2 n+1)+(n+1)^2 =\\
(n+1) \left(\frac{1}{6} n (2 n+1)+(n+1) \right)=\\
\frac{1}{6} (n+1) (n+2) (2 (n+1)+1)

\]

Answer 6

As an exercise, show that
\[
\sum _{i=0}^n i^3=\frac{1}{4} n^2 (n+1)^2
\]

Answer 7

Beautiful! I will do that!
I'm trying to show that \[ \sum_{i=0}^{n+1}i^3 = \frac{1}{4}(n+1)^2(n+2)^2 \] by induction so:

\[ \sum_{i=0}^{n+1}i^3 = \frac{1}{4}n^2(n+1)^2 + (n+1)^3 \\ = (n+1)^2(\frac{1}{4}n^2+n+1) \\ = \frac{1}{4}(n+1)^2(n^2+4n+4) \\ = \frac{1}{4}(n+1)^2(n+2)^2 \]

Woo! That was fun. I can now show they are equal by induction, but I still would like to learn how to find the equation that gives the n term of the series. That may be out of the scope of this question, but is there a certain method for doing this or does one learn simply by experience? I am going to read the wiki article http://en.wikipedia.org/wiki/Mathematical_induction.

Thank you a bunch, @eliassaab!!!