1.  Take the sum  of all integer numbers multiple of 3 from 1 to 4133.
    It is the sum of an AP with the first term 3, the common difference 3 and the last term of 4131. 

    The number of terms is n =  = 1377.


2.  Take the sum  of all integer numbers multiple of 5 from 1 to 4133.
    It is the sum of an AP with the first term 5, the common difference 5 and the last term of 4130. 

    The number of terms is n =  = 826.


3.  Calculate .

    Still, it is NOT your answer.


4.  To get the answer, distract from  the sum  of all integer numbers multiple of 15 from 1 to 4133. 

         ( Since you counted these numbers TWICE in the sum  (!) )

    The sum  is the sum of an AP with the first term 15, the common difference of 15 and the last term 4125.

    The number of terms is n =  = 275.


5.  Again, your answer will be .

    Do all these calculations on your own.

    You have clear instructions. The rest is simple arithmetic.

If we find the sum of all the multiples of 3 below 4133,
and add that to the sum of all the multiples of 5 below 4133,
we will have added the multiples of 15 below 4133 twice.  So 
then if we subtract the sum of all the multiples of 15 from
that we'll have the desired sum.

We find the sum of the multiples of 3 below 4133:

The least multiple of 3 below 4133 is 3.

4133/3 = 1377.67 so the largest multiple of 3 less that 4133
is 1377*3 or 4131. 

We want the sum of the arithmetic sequence:  

3,6,9,...,4131

We use the sum formula



We need a1=3 and an=4131

It's obvious to me that the number of terms is n = 1377, but if
it's not obvious to you, just divide them all by 3 and
get the sequence

1,2,3,...,1377

and now you can see that indeed n = 1377.

We substitute in the sum formula



--------------------

We find the sum of the multiples of 5 below 4133:

The least multiple of 5 below 4133 is 5.

4133/5 = 826.6 so the largest multiple of 3 less that 4133
is 826*5 or 4130. 

We want the sum of the arithmetic sequence:  

5,10,15,...,4130

We use the sum formula again:



We need a1=5 and an=4130

As before, it's obvious to me that the number of terms is 
n = 826, but again, if it's not obvious to you, just divide 
them all by 5 and get the sequence

1,2,3,...,826

and now you can see that indeed n=826.

We substitute in the sum formula



---------------------------

We find the sum of the multiples of 15 below 4133:

The least multiple of 15 below 4133 is 15.

4133/15 = 275.33 so the largest multiple of 15 less that 4133
is 275*15 or 4125. 

We want the sum of the arithmetic sequence:  

15,30,45,...,4125

We use the sum formula



We need a1=15 and an=4125

Once more, it's obvious to me that the number of terms is 
n = 275, but if it's not obvious to you, just divide them all 
by 15 and get the sequence

1,2,3,...,275

and now you can see that indeed n=275.

We substitute in the sum formula



-----------------------------------------------

So we add the sum of the multiples of 3 and the sum of the
multiples of 5, and get 2846259+1707755 = 4554014

That adds the multiples of 15 twice, since multiples of 15
are multiples of both 3 and 5.

So we subtract the sum of the multiples of 15 and get

4554014-569250 = 3984764  <-- answer

Edwin