f(x)=x2+x,则数列{1/fn}的前n项和为多少?我一年多没做过数列题了,

f(x)=x2+x,则数列{1/fn}的前n项和为多少?我一年多没做过数列题了,
f(x)=x2+x,则数列{1/fn}的前n项和为多少?
我一年多没做过数列题了,

f(x)=x2+x,则数列{1/fn}的前n项和为多少?我一年多没做过数列题了,
利用差分法
1/fn=1/(n²+n)=1/(n*(n+1))=1n-1/(n+1)
所以前n项和
Sn=(1-1/2)+(1/2-1/3)+...+(1/n-1/(n+1))
=1-1/(n+1)=n/(n+1)

f(n)=n²+n=n*(n+1)
1/f(n)=1/[n*(n+1)]=1/n-1/(n+1)
Sn=f(1)+f(2)+f(3)+······f(n)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+······[1/n-1/(n+1)]
=1-1/2+1/2-1/3+1/3-1/4+······1/n-1/(n+1)
=1-1/(n+1)=n/(n+1)