Question

高中数学等差数列等比数列问题

Answer 1

an=a1+(n-1)d
a5=2
a1+4d=2 (1)
a6+a7+a8=24
3a1+18d=24
a1+6d=8 (2)
(2)-(1)
2d=6
d=3
from (1)
a1+4d=2
a1=-10
an = -10+3(n-1) = 3n -13
(2)
Sn = a1+a2+...+an = (a1+an)n/2 = n(3n -23)/2
bn.(2Sn+26n) =1

bn.(n(3n -23)+26n) =1
bn.(3n^2+3n) = 1
bn = 1/[3n(n+1)]
=(1/3) [ 1/n -1/(n+1) ]
b1+b2+..+bn = (1/3) [ 1 - 1/(n+1) ] = n/[3(n+1)]
(3)
cn
=(an+n+12).3^n
=( 3n -13 +n + 12) .3^n
=(4n -1 ) .3^n
=4( n.3^n) - 3^n
let
S= 1.3^1+2.3^2+....+n.3^n (3)
3S= 1.3^2+2.3^3+....+n.3^(n+1) (4)
(4)-(3)
2S =n.3^(n+1) - ( 3^1+3^2+...+3^n)
=n.3^(n+1) -(3/2) ( 3^n -1)
4S =2n.3^(n+1) -3( 3^n -1)
Tn
= c1+c2+...+cn
= 4S - (3/2)(3^n -1 )
=2n.3^(n+1) -3( 3^n -1) -(3/2)(3^n -1 )
=2n.3^(n+1) -(9/2)( 3^n -1)
=9/2 + (6n - 9/2) .3^n