Compute \([\sum_{n=1}^{1000} \frac{1}{n^2 + n}]\)
+128472 Note that we can write 1 / [ n^2 + n] as 1/ n - 1 / [ n + 1 ]
So we can write this series in this manner :
[ 1/1 - 1/ 2] + [ 1/2 - 1/3] + [ 1/3 - 1/4] + [ 1/4 - 1/5] + ...... + [ 1/998 - 1/999] + [ 1/999 - 1/1000] + (1/1000 - 1/1001) =
1 + (1/2 - 1/2) + (1/3 - 1/3) + (1/4 - 1/4) + .......+ (1/998 - 1/998) + (1/999 - 1/999) - (1/1000 -1/1000) - 1/1001 =
1 - 1/1001 =
[ 1001 - 1 ] / 1001 =
1000 / 1001
