how do I simplify $1+4+9+\cdot\cdot\cdot+x^2$

Question

zzr0ck3r · Answer

$\sum_{i=1}^ni^2=1+4+9+16+..+n^2$

kainui · Answer

Whether this is what you want or not, oh well I enjoy this too much to stop from writing it out.

Take note of this telescoping series:

$$\sum_{k=1}^x k^3 - (k-1)^3 = [1^3-0^3]+[2^3-1^3]+\cdots + [x^3 - (x-1)^3]= x^3$$
Now you can expand the inner sum:
$$\sum_{k=1}^x k^3-(k-1)^3 = x^3$$$$\sum_{k=1}^x k^3 - (k^3-3k^2+3k-1) =x^3$$$$\sum_{k=1}^x 3k^2-3k+1 = x^3$$

You can pick apart this sum of 3 terms into 3 sums:
$$3\sum_{k=1}^x k^2-3\sum_{k=1}^x k+\sum_{k=1}^x1 = x^3$$
The first sum is what we want, the second sum you might recognize from Gauss's trick (one of his many tricks really lol) and the last sum is just counting.
$$\sum_{k=1}^xk = \frac{x(x+1)}{2}$$$$\sum_{k=1}^x 1 = x$$
Throw those in to get:

$$3\sum_{k=1}^x k^2-3\sum_{k=1}^x k+\sum_{k=1}^x1 = x^3$$
$$3\sum_{k=1}^x k^2-3\frac{x(x+1)}{2}+x = x^3$$
Little algebra away now:
$$\sum_{k=1}^x k^2 = \frac{x^3}{3} + \frac{x(x+1)}{2} - \frac{x}{3}$$
Little more tidied up:
$$\sum_{k=1}^x k^2 = \frac{x(x+1)(2x+1)}{6}$$

Hopefuly I didn't make any mistakes, if I did pls forgive the idea is still in tact I hope. The point is you can easily(heh) make a telescoping series, and from that you can get an expression for any sum of powers.