已知数列{an}中 an=1/n^2+n 则前100项的和为多少

an=1/n^2+n=1/n*(n+1)=1/n-1/(n+1)

S=1/(1^2+1)+1/(2^2+2)+..+1/(100^2+100)
=1/1*2+1/2*3+..+1/100*101
=1/1-1/2+1/2-1/3+...+1/100-1/101
=1-1/101
=100/101

先分解因式,再裂项

1/(n^2+n)=(1/n-1/(n+1)),故
则前100项的和为1/1-1/101=100/101

an=(1/n-1/n+2)/2

Sn=(1/1+1/2-1/n+1-1/n+2)/2

s100=请自己算之