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数列求和高考题_数列求和高考题20道
tamoadmin 2024-05-18 人已围观
简介(1)2Sn = 3^n+3n=1,2a1=3+3a1=3for n>=2an = Sn - S(n-1)2an = 3^n - 3^(n-1)an = 3^(n-1)(2)letS= 1.(1/3)^0 +2.(1/3)^1+....+n.(1/3)^(n-1) (1)(1/3)S= 1.(1/3)^1 +2.(1/3)^2+....+n.(
(1)
2Sn = 3^n+3
n=1,
2a1=3+3
a1=3
for n>=2
an = Sn - S(n-1)
2an = 3^n - 3^(n-1)
an = 3^(n-1)
(2)
let
S= 1.(1/3)^0 +2.(1/3)^1+....+n.(1/3)^(n-1) (1)
(1/3)S= 1.(1/3)^1 +2.(1/3)^2+....+n.(1/3)^n (2)
(1)-(2)
(2/3)S = [1+1/3+...+1/3^(n-1) ]- n.(1/3)^n
= (3/2)( 1- 1/3^n) - n.(1/3)^n
S = (9/4)( 1- 1/3^n) - (3/2)n.(1/3)^n
an.bn= log<2>an
bn = (n-1) log<2>3/ 3^(n-1)
= (log<2>3) . [ n.(1/3^(n-1)) -1/3^(n-1) ]
Tn = b1+b2+...+bb
= (log<2>3) . [ S - (3/2)(1- 1/3^n) ]
= (log<2>3) . [ (9/4)( 1- 1/3^n) - (3/2)n.(1/3)^n - (3/2)(1- 1/3^n) ]
= (log<2>3) . [ (9/4)( 1- 1/3^n) - 3n.(1/3)^n ]
= (log<2>3) .(1/4) [ 9( 1- 1/3^n) - 12n.(1/3)^n ]
= (log<2>3) .(1/4) [ 9-(12n+9) (1/3)^n ]
