Another summation

The set of whole numbers starts from 0 and consists of all positive integers.

So, the series of the first 2014 whole nos. is : $0,1,2,3,.....,2013$ . Now, we use here the identity $1+2+3+....+n=\frac{n(n+1)}{2}$

$0+1+2+3+...+2013 = 1+2+3+....+2013 = \frac{2013\times 2014}{2}=2013\times 1007 = \boxed{2027091}$

So, the last three digits of the sum is $=\boxed{091}$

Also, you can find the last three digits of the sum by multiplying the numbers formed by the last three digits of $1007$ and $2013$ , i.e, by multiplying the numbers $007$ and $013$ and checking the last three digits of that product.

$007\times 013 = 7\times 13 = 91 = 091$

So, the last three digits of that product is $091$ , so the last three digits of the original product is also $\boxed{091}$