Question

数列问题求大神解答

Answer 1

收

Answer 2

你就不能先搜题了有不会的再问么。如果搜不到了就问你哪不会，除非你全都不会

追问

搜不到

Answer 3

(1)
ana(n+1)+4=4an
a(n+1)=(4an-4)/an
2/[a(n+1)-2]=2/[(4an-4)/an -2]
=an/(an-2)
=(an -2+2)/(an-2)
=2/(an-2) +1
2/[a(n+1) -2]- 2/(an-2)=1，为定值
2/(a1-2)=2/(4-2)=1
数列{2/(an-2)}是以1为首项，1为公差的等差数列
(2)
2/(an -2)=1+1·(n-1)
an-2=2/n
bn=(an-2)[a(n+1)-2]
=(2/n)[2/(n+1)]
=4[1/n -1/(n+1)]
Tn=4[1/1 -1/2 +1/2 -1/3+...+1/n -1/(n+1)]
=4[1- 1/(n+1)]
=4n/(n+1)

Answer 4

如图