Question

数学小题

通项公式为1/（n^2+n)的数列，其Sn如何求？
请高人讲解一下如何用递推公式得其Sn？

Answer 1

1/(n^2+n)=1/n-1/(n+1)
Sn=1-1/2+1/2-1/3+1/3-1/4+……+1/n-1/(n+1)约去相同的，得Sn=1-1/（n+1）

Answer 2

1/（n^2+n)=1/n - 1/(n+1)
然后递推

Sn=1-1/2+1/2-1/3+...+1/n - 1/(n+1)
=1-1/(n+1)
=n/(n+1)

Answer 3

Sn=1/（n^2+n)= 1/n*(n+1)

n=1, S1=1/2
n=2, S2=1/6
....
n=100, S100=100*101=101000

Answer 4

an=1/（n^2+n)=1/n-1/(n+1)

Sn=1-(1/2)+(1/2)-(1/3)+…………+1/(n-1)-1/n+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)