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 =5^{2}. If we had a square of dimensions n by n, it would contain n^{2} pentagons.

Now simply put it all together. The total number of pentagons is n^{2} but it’s also equal to the sum of reds(S) and blues (S-n): ** n ^{2} = S + S – n**

Solve for S:

2 S = **n ^{2}+ n**

S = ( **n ^{2}+ 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 (n^{2} + 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!

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