# How they get to this formula

1. Mar 13, 2009

### transgalactic

$$1^2+2^2+..+k^2=\frac{k(k+1)(k+2)}{6}$$

how to get this general result
??

i know

1+2+..+n=\frac{n(n+1)}{2}

so by that rule the sum should be
$$\frac{n(n^2+1)}{2}$$

2. Mar 13, 2009

Two methods come quickly to mind.

First, since you are given that the sum should equal

$$\frac{k (k+1)(k+2)}{6}$$

prove the result by induction.

Alternatively, in general the sum

$$1 + 2^r + 3^r + \dots + k^r$$

for integer $$r$$ is a polynomial of degree $$r + 1$$. For $$r = 2$$

$$1 + 2^2 + 3^2 + \dots + k^2 = ak^3 + bk^2 + ck + d$$

Evaluate the left side for four different values of $$k$$ (0, 1, 2, 3) and solve the system of linear equations for the coefficients.

3. Mar 13, 2009

### sutupidmath

Or you can use the teleschoping series, to prove that.

i.e

$$\sum_{i=1}^{k}i^2=\frac{k(k+1)(k+2)}{6}$$

4. Mar 13, 2009

### transgalactic

i cant understand how you get the equations out of you second alternative methos

5. Mar 13, 2009

### D H

Staff Emeritus
You can't, because it's not true. Try it with k=2. 12+22=5, 2*3*4/6=4.

What makes you think that? Hint: It's not valid either.

6. Mar 13, 2009

### Mentallic

Thanks, I found this very interesting

Through the method shown by statdad, this sum is actually:

(pending for you to find out)

Last edited: Mar 13, 2009
7. Mar 13, 2009

### D H

Staff Emeritus
Bingo! But you shouldn't give answers away.