Question 723119
That is the sum of an arithmetic sequence with 1000 terms.
It is easy to see that the sum,
4+8+12+ .... +3992+3996+4000 is half of
(4+8+12+ .... +3992+3996+4000) + (4000+3996+3992+ .... +12+8+4) = (4+4000)+(8+3996)+(12+3992)+ ... +(3992+12)+(3996+8)+(4000+4)=4004+4004+4004+ ... +4004+4004+4004=1000*4004=4004000
So the sum is {{{highlight(2002000)}}} .
 
We could write some fancy formulas, like
{{{a[1]=4}}}, {{{d=4}}}, {{{a[k]+4=a[k+1]}}},
{{{a[k]=a[1]+d(k-1)}}} and in this case {{{a[k]=4+4(k-1)=4+4k-4=4k}}}
In this case, {{{n=1000}}} {{{a[n]=4*1000=4000}}}
{{{sum(a[k],k=1,1000)=n*(a[1]+a[n])/2}}} and in this case
{{{sum(4k,k=1,1000)=1000*(4+4000)/2=1000*4004/2=2002000}}}