等差数列bn=(a1+a2+3a….an)/n(1)bn=n^2,求｛an｝(2){bn}为等差数列,求证｛an｝也为等差数列错了_数学解答

等差数列bn=(a1+a2+3a….an)/n
(1)bn=n^2,求｛an｝
(2){bn}为等差数列,求证｛an｝也为等差数列
错了是（a1+a2-+a3...+an）/n

人气：344 ℃　时间：2020-06-19 16:18:02

解答

bn=(a1+a2+3a….an)/n=Sn/nb1=a1bn=n^2,a1=b1=1sn=n^3s(n+1)=(n+1)^3a(n+1)=s(n+1)-sn=3n^2+3n+1=3n(n+1)+1所以an=3n(n-1)+1 n>=2当n=1时,a1=1；2）如果bn是等差数列,不妨设bn=kn+d；则Sn=（kn+d）ns（n+1）=（kn+k+...