Question

find the sum of all natural numbers which are multiples of 2 and 3 less than 301.​

Answer 1

Answer:

∑k=1nk=12n(n+1)

Step-by-step explanation:

The previously posted answer isn't correct. The statement of the problem is to sum the multiples of 3 and 5 below 1000, not up to and equal 1000. The correct answer is

∑k1=13333k1+∑k2=11995k2−∑k3=16615k3=166833+99500−33165=233168,

where we have the used the identity

∑k=1nk=12n(n+1).