Question

如图，数列问题。

Answer 1

1)
na(n+1)=(n+1)an+1
等式两边除以n(n+1)得
a(n+1)/(n+1)=an/n+[1/n(n+1)]
[a(n+1)/(n+1)]-an/n=1/n(n+1)
[a(n+1)/(n+1)]-an/n=(1/n)-[1/(n+1)]
累加得:
a(n+1)/(n+1)]-a1=[(a2/2)-a1]+[(a3/3)-a2/2]+...+[a(n+1)/(n+1)]-an/n
=(1/-1/2)+(1/2-1/3)+...+(1/n)-[1/(n+1)]=1-[1/(n+1)]=n/(n+1)
所以有:a(n+1)/(n+1)]-a1=n/(n+1)
a(n+1)/(n+1)]=1+[n/(n+1)]
a(n+1)/(n+1)]=(2n+1)/(n+1)
a(n+1)=(2n+1)
an=2n-1