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Proof that the sum of the squares of the first n whole numbers is `n^3/3 + n^2/2 + n/6`

A recent thread on Hacker News that I started with a flippant comment turned into a little mathematical puzzle.

What's the sum of the square of the first n whole numbers?

It's well known that the sum of the first

It turns out that it's easy to prove that

If we fill in the blank squares to make a rectangle we have the basis of a derivation of the formula:

Looking at the formerly blank squares (that I've numbered to assist with the thinking) we can see that the columns have

Now the total rectangle is

So we can calculate

What's the sum of the square of the first n whole numbers?

It's well known that the sum of the first

`n`whole numbers is`n(n+1)/2`. But what's the value of`sum(i=1..n) n^2`? (I'll call this number`S`for the remainder of this post).It turns out that it's easy to prove that

`S = n^3/3 + n^2/2 + n/6`by induction. But how is the formula derived? To help with reasoning here's a little picture of the first 4 squares stacked up one on top of the other:If we fill in the blank squares to make a rectangle we have the basis of a derivation of the formula:

Looking at the formerly blank squares (that I've numbered to assist with the thinking) we can see that the columns have

`1`then`1+2`then`1+2+3`and finally`1+2+3+4`squares. Thus the columns are sums of consecutive whole numbers (for which we already have the`n(n+1)/2`formula.Now the total rectangle is

`n+1`squares wide (in this case`5`) and its height is the final sum of whole numbers up to`n`or`n(n+1)/2`(in the image it's`4 x 5 / 2 = 10`. So the total number of squares in the rectangle is`(n+1)n(n+1)/2`(in the example that's`5 x 10 = 50`).So we can calculate

`S`as the total rectangle minus the formerly blank squares which gives:

S = (n+1)n(n+1)/2 - sum(i=1..n)sum(j=1..i) j

= (n(n+1)^2)/2 - sum(i=1..n) i(i+1)/2

2S = n(n+1)^2 - sum(i=1..n) i(i+1)

= n(n+1)^2 - sum(i=1..n) i^2 - sum(i=1..n) i

= n(n+1)^2 - S - n(n+1)/2

3S = n(n+1)^2 - n(n+1)/2

= n(n+1)( n+1 - 1/2 )

= n(n+1)(n+1/2)

= (n^2+n)(n+1/2)

= n^3 + n^2/2 + n^2 + n/2

= n^3 + 3n^2/2 + n/2

S = n^3/3 + n^2/2 + n/6

Labels: mathematics

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