已知数列｛an｝满足a1=1/2,a（n+1）=an+1/（n平方+n）,求an

题目

已知数列｛an｝满足a1=1/2,a（n+1）=an+1/（n平方+n）,求an

答案

a(n+1)=an+1/[n(n+1)]
=>a(n+1)-an=1/[n(n+1)]=1/n-1/(n+1)
当n≥1时,
a2-a1=1-1/2
a3-a2=1/2-1/3
a4-a3=1/3-1/4
...
an-a(n-1)=1/(n-1)-1/n
a(n+1)-an=1/n-1/(n+1)
将上述等式两边分别相加,则有
(a2-a1)+(a3-a2)+(a4-a3)+...+[an-a(n-1)]+[a(n+1)-an]=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+[1/(n-1)-1/n]+[1/n-1/(n+1)]
=>a(n+1)-a1=1-1/(n+1)
=>a(n+1)=3/2-1/(n+1)
=>an=3/2-1/n