In the previous essay, we saw how 1 + 2 + 3 +… n = n(n+1)/2, which can also be expressed using the sum symbol as

But what if we wanted a formula for the sum of sums, specifically , 1 + 3 + 6 + 10 + … n(n+1)/2, Notice that 1 = 1; 3 = 1+ 2; the 3rd term = 6 = 1+ 2 + 3; the 4th term =10 = 1 + 2 + 3 + 4, and all the way up to n(n+1)/2. How do we get a formula for such a sum of sums?

To derive a formula for this sum S, use our n(n+1)/2 formula to express each number in the sum 1 + 3 + 6 + 10 + … n(n+1)/2 as

We notice that S is half the sum of two different sums: the sum of squares up to n and the sum of numbers up to n. We will call the first sum W, and the second sum is the familiar n(n+1)/2

therefore

So now all we have to do is get an expression for W and then we will be able to get S in terms of n.

Consider the following:

The sum of black squares is 1+ 3 + 6 + 10, but if we imagine them going up to n(n+1)/2, the sum of sums, we have essentially the sum that we were looking for S.

So the total area of our rectangle, R, is really just equal to W + S.

R = W + S

But R is also equal to its length, n(n+1)/2, multiplied by its width, n+1:

We isolate the W :

Recall:

Substitute our expression for W into the above and we obtain:

For example if we choose n = 8, the sum of sums , S = 1 + 3 + 6 + 10 +… 8(8+1)/2=

1 + 3 + 6 + 10 +15 + 21+ 28 + 36 = 8(9)(10)/6 =120 . Add the components of the sequence the long way, and the sum is indeed 120.

Now that we have S, we can also obtain a formula for the sum of squares, W. We’ve seen that

Substitute for S, simplify and factor:

For example 1^{2}+ 2^{2} + 3^{2} + 4^{2} + 5^{2} = 55, but we could use our formula:

Gong back to our sum notation we can summarize what we have found using a little algebra, geometry and, most importantly, a little creativity.

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