# deriving formula for sum of squares?

How would you derive a formula for the sum of squares from 1 to n. such that f(n) = 1^2 + 2^2 + 3^2 ...... n^2. I've worked out the formula (its a polynomial), but I would like a neat way of deriving it.

### 8 Answers

- 1 decade agoFavorite Answer
As n increases, f(n) increases at a rate proportional to n^2. Therefore, since the rate of change of f(n) is quadratic, f(n) must be cubic. Therefore it is of the form f(n)=an^3+bn^2+cn+d. Now simply take four values from your formula, plug them into the equation, and solve for the constants.

f(1)=1=a+b+c+d

f(2)=5=8a+4b+2c+d

f(3)=14=27a+9b+3c+d

f(4)=30=64a+16b+4c+d

After you solve this system of equations you get a=1/3, b=1/2, c=1/6, and d=0. So, f(n)=x^3/3+x^2/2+x/6. This formula works for values of n from 1 to 4, thus confirming that my work is right. It also works for n=5 (f(n)=55) and for n=6 (f(n)=91), suggesting that it is true for all n.

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- mathematicianLv 71 decade ago
There are several ways of doing this. One that I like is to notice that the sum of

(k+1)^3 -k^3=

3k^2 +3k +1

from k=1 to k=n is (n+1)^3 -1^3 (the sum telescopes).

You know the sum of k from 1 to n and the sum of 1 from 1 to n. This allows the solution of the sum of k^2 from 1 to n:

sum k^2 =

=(1/3)[(n+1)^3 -1^3 -3*sum k -sum 1]

=(1/3)[n^3 +3n^2 +3n -(3/2)(n^2 +n) - n]

=(1/3)n^3 +(1/2)n^2 +(1/6)n.

The same technique can be used to find sums of higher powers.

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- bubsirLv 41 decade ago
Here's a way, I don't know how "clever" it is:

n^2____= n + n + n + ... + n + n n times

(n-1)^2 = n-1 + n-1 + n-1 + ... + n-1 n-1 times

.

(n-k)^2 = n-k + n-k + n-k + ... + n-k n-k times

2^2 ___= 2 + 2

1^2 ___= 1

Sum the columns (sorry they don't line up too well in this editor)

noting that sum from n to k is (k^2-n^2+k+n)/2

so sum from 1 to k on n of (k^2-n^2+k+n)/2

is going to be sum(n^2).

Set them equal to eachother and solve.

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- cassieLv 61 decade ago
Ok clever clogs if you can work out the formula, you can work out a way of deriving it me thinks.

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- 1 decade ago
This is not my site, but a fairly concise derivation of the formula is here:

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- LearnerLv 41 decade ago
I wouldn't have an idea....but would just LOVE to be as clever as all you people. I try to grasp the whole thing here and read the Mathematics for enjoyment.Well done all of you.

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- Anonymous1 decade ago
Look at the sum of cubes, its mad as fook, its the sum of integers squared.

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- 1 decade ago
These types of formulas can all be proven by induction.

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