来自兰中文的问题
a1=1an=an-1加1/n(n+3)求ann-1为角标,
a1=1an=an-1加1/n(n+3)求an
n-1为角标,
1回答
2020-08-10 07:55
a1=1an=an-1加1/n(n+3)求ann-1为角标,
a1=1an=an-1加1/n(n+3)求an
n-1为角标,
an=an-1+1/n(n+3)
=a(n-1)+(1/n-1/(n+3))/3
=a(n-2)+(1/(n-1)-1/(n+2))/3+(1/n-1/(n+3))/3
=...
=a1+(1/2-1/5)/3+(1/3-1/6)/3+...+(1/n-1/(n+3))/3
=1+(1/2+1/3+1/4-1/(n+1)-1/(n+2)-1/(n+3))/3
=1+13/36-(1/(n+1)+1/(n+2)+1/(n+3))/3
=49/36-(1/(n+1)+1/(n+2)+1/(n+3))/3
验算一下a1=1
所以应为正an=49/36-(1/(n+1)+1/(n+2)+1/(n+3))/3
如仍有疑惑,欢迎追问.祝: