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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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    @GerryMyerson that explains it.2012-08-03

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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.