# Sum of mutliples b/w $2$ and $10$ . What is wrong with this method

I am trying to find the sum of multiples b/w $2$ and $10$ end points non-inclusive using the following mechanism and I don't know why I am getting the wrong answer. (I selected this small range intentionally just to check the result)

First I find the no. of multiples b/w $2$ and $10$ using

No. of Multiples = $\frac{10-2}{2} - 1 = 3$ So there are three multiples of 2 b/w $2$ and $10$

Sum of Multiples now is = Multiple$(1+2+3)$=$2(6)=12$

But The answer is supposes to be $4+6+8 = 18$ Why am I getting the wrong answer.

Edit: I think the problem is with the formula:

Sum of Multiples now is = Multiple$(1+2+3)$=$2(6)=12$

I got this from here 3rd Comment after the question

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Yes. I am trying to find the sum of multiples of 2 between $2$ and $10$. These multiples are 4,6 and 8. And their sum is 4+6+8 = 18 –  MistyD Aug 3 '12 at 3:44
I take it you're trying to find the sum of the multiples of 2 between 2 and 10, endpoints excluded. Note the right answer is $2(2+3+4)$; where did you get your $1+2+3$? –  Gerry Myerson Aug 3 '12 at 3:44
@GerryMyerson Yes thats what I am trying to do. I thought that we were supposed to add till the required no of multiples ? –  MistyD Aug 3 '12 at 3:46
Link Have a look at the 3rd comment. He started from 1 and went uptil the number of the multiples –  MistyD Aug 3 '12 at 3:49
Yes, but in that question the endpoint was included, so the first number being summed was 1 times the common multiplier; you're starting with 4, not 2, and 4 isn't $2\times1$, it's $2\times2$. –  Gerry Myerson Aug 3 '12 at 3:54

Maybe things will be clearer if we look at another example. Find the sum of all the multiples of 3 between 10 and 100. In this case, we are being asked to find $$12+15+18+21+\cdots+93+96+99.$$ That's the same as $$3(4+5+6+7+\cdots+31+32+33).$$ Why do the numbers inside the bracket start at 4, not at 1? Because the numbers we're adding start at 12, which is $3\times4$, not $3\times1$. Why do the numbers inside the bracket go up to 33? Because the numbers we're adding go up to 99, and $99=3\times33$.
This still leaves you with the problem of finding $$4+5+6+7+\cdots+31+32+33,$$ of course.