The sum of an arithmetic progression is $$S_n=\frac{n*(a_1+a_n)}{2}$$. In our case $$a_n = a_1 * n$$ so we have $$S_n=\frac{a_1*n*(n+1)}{2}$$. We use the above formula to calculate the sum of the multiplies of 3 and 5 minus the sum of the multiplies of 15 (since $$3 * 5 = 15$$, multiplies of 15 are counted two times).