Jose has a project that requires him to find the sum of the multiples of three from 3 all the way to 999, including those two numbers. If he uses the small number plus large number method, what is the sum of the multiples of three from 3-999? Note that he will also need to know how many numbers has in order to use that method.

The sum Jose wants to compute is

3 + 6 + 9 + … + 993 + 996 + 999

but since each number is a multiple of 3, he can remove a factor of 3 from eac term:

3 (1 + 2 + 3 + … + 331 + 332 + 333)

Then it's clear that the sum contains 333 terms.

Let S be the value of the sum of the first 333 natural numbers:

S = 1 + 2 + 3 + … + 331 + 332 + 333

Reversing the order of terms has no effect on this value:

S = 333 + 332 + 331 + … + 3 + 2 + 1

Now notice that terms in the same position sum to 334:

1 + 333 = 334

2 + 332 = 334

3 + 331 = 334

and so on. This means that if we pair up terms in the positions and add them up, all we are doing is doubling the value of S on the left side, and adding up 333 copies of 334 on the right side:

2S = 334 + 334 + … + 334 + 334

2S = 333 × 334

Solve for S :

S = 333 × 334 / 2 = 55,611

Then the sum Jose is considering is 3 times this value,

3 + 6 + 9 + … + 993 + 996 + 999 = 166,833

step-by-step explanation:

a rhombus is a parallelogram enjoying his symmetry.