# The Sum of Sums

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

S= 1(1+1)/2 + 2(2+1)/2 +3(3+1)/2 +4(4+1)/2 +….n(n+1)/2

S= (1/2)( 1(1+1) + 2(2+1) +3(3+1) +4(4+1) +….n(n+1)

Expand:

S= (1/2)( 12+ 1 + 22 + 2 + 32 + 3 + 42 + 4 +…n2 + n )

Group the squares together :

S= (1/2)( 12+ 22 + 32 + 42 +…n2 + 1 + 2 + 3 + 4 + …. n )

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 12+ 22 + 32 + 42 + 52 = 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.

Postscript: To modify the formula so that the sum of sums, S, is in terms of the last term, k:

# From Dominoes to Math Summation Formulas

Standard dominoes have 28 tiles. Figuring why is fairly straightforward. Think of all the tiles with a blank. There are seven in all.

When we start with 2, the blank-2 and 1-2 combination are already counted, so we are left with only 5 options, pairing 2 with the numbers 2 through 6.

The pattern is clear; the total number of dominoes will be 7+6+5+4+3+2+1 =28.

But what if we had dominoes that went up to 1000? How many would there be? Taking the blank into consideration, is there a quick way of obtaining the sum 1 + 2 + 3+…1001? Legend has it that Gauss figured this out when he was a little kid in elementary school. Regardless of the story’s veracity, we don’t have to be Gaussian geniuses to figure it out.

Consider the following:

If we look at the diagonals, the total number of red pentagons is represented by the sum 1 + 2 + 3 + 4 + 5. But let’s imagine that the largest number was not 5 but something much bigger, which we will call n. The sum(S) of red pentagons would then be 1+ 2+ 3…+ n = S.

Notice that the sum of blue pentagons is missing the n row, so that in all, there are S-n blue pentagons. In reality we have 25 pentagons pictured; 25 =52. If we had a square of dimensions n by n, it would contain n2 pentagons.

Now simply put it all together. The total number of pentagons is n2 but it’s also equal to the sum of reds(S) and blues (S-n): n2 = S + S – n

Solve for S:

2 S = n2+ n

S = ( n2+ n )/2, and if you factor the expression

S = n( n+1 )/2.

In other words, to quickly obtain the sum of regular dominoes, we could have taken half of 8 and multiplied by 7, which equals 28. If we wanted the sum up to 1001, it would simply have been

1001(1002/2) = 501 501. ( incidentally, any time you multiply a 3-digit number by 1001, the three digits are repeated. )

What if we wanted to add up the even numbers up to n?

2+ 4 + 6 + 8 +…n = E.

Simply factor out the 2:

2(1 + 2 + 3 +…n/2) = E.

2 times the Sum up to n/2 =E = sum of even numbers up to the even number n.

In brackets we have the sum of consecutive integers up to n/2. So replace the n from the S formula with n/2:

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

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

or (n2 + 2n)/4 + E

factoring we obtain:

n(n+2)/4 = E

Try it 2 + 4+ 6+ 8 + 10 should be 30 = 10(12)/4 . It works!

What about the sum of odd numbers up to the odd number n?

D = Sum of all numbers up to n minus the sum of even numbers up to n-1( one less than the odd number n)

D = n( n+1 )/2 – (n-1)(n-1+2)/4.

Notice that in the even number part of the formula, we had to replace n with n-1.

Simplifying we get :

D = (n+1)2/4

Test it for 1+ 3 + 5 + 7 + 9 + 11 = (11+1)2/4 =36, which again is correct!