第2个回答 2020-05-28
∵a1+2a2+3a3+···+nan
=(n+1)/2·a(n+1),①
∴a1+2a2+3a3+···+(n-1)a(n-1)
=n/2·an②
由①-②得:
nan=(n+1)/2·a(n+1)-n/2·an,
∴3n/an=(n+1)/2·a(n+1),
∴a(n+1)/an=3·n/(n+1),
∴an/a(n-1)=3·(n-1)/n,
····
a3/a2=3·(3+1)/3,
a2/a1=3·(2+1)/2,
累乘得:an/a1=3^(n-1)/n,
又∵a1=1,
所以所求为:an=3^(n-1)/n。本回答被网友采纳