数列bn=2^n(4n-3),求Sn
来源：学生作业帮助网 编辑：六六作业网 时间：2025/01/09 17:40:02
数列bn=2^n(4n-3),求Sn
数列bn=2^n(4n-3),求Sn
数列bn=2^n(4n-3),求Sn
bn=2^n(4n-3)
= 8[n(2^(n-1))] - 3.2^n
consider
(x^(n+1)-1)/(x-1) = 1+x+x^2+..+x^(n)
[(x^(n+1)-1)/(x-1)]' = 1+2x+3x^2+...+nx^(n-1)
1+2x+3x^2+...+nx^(n-1) = [(x-1)((n+1)x^n- (x^(n+1)-1)]/(x-1)^2
= [( nx^(n+1) -(n+1)x^n+1]/(x-1)^2
put x=2
1+2(2)+3(2)^2+...+n(2^(n-1)) = (n)2^(n+1)-(n+1)2^n +1
bn=2^n(4n-3)
= 8[n(2^(n-1))] - 3.2^n
summation bn
= 8[(n)2^(n+1)-(n+1)2^n +1] - 6(2^n-1)
= 8[(n)2^(n+1)-(n+1)2^n +1] - 3(2^(n+1)) +6
=(8n-3).2^(n+1) - 8(n+1)2^n +14
错位相减法
Sn=1×2+5×2²+9×2^3+...........+(4n-3)*2^n (1)
2Sn=2²+5×2^3+9×2^4+.........+(4n-7)*2^n+(4n-3)*2^(n+1) (2)
(1)-(2):
-Sn=2+4×2^2+4×2^3+..........+4×2^n-(...
全部展开
错位相减法
Sn=1×2+5×2²+9×2^3+...........+(4n-3)*2^n (1)
2Sn=2²+5×2^3+9×2^4+.........+(4n-7)*2^n+(4n-3)*2^(n+1) (2)
(1)-(2):
-Sn=2+4×2^2+4×2^3+..........+4×2^n-(4n-3)×2^(n+1)
=2+16[2^(n-1)-1]/(4-1)-(4n-3)*2^(n+1)
=2+4/3*2^(n+1)-16/3-(4n-3)*2^(n+1)
=-10/3-(4n-13/3)*2^(n+1)
∴Sn=10/3+2^(n+1)(12n-13)/3
=[10+2^(n+1)*(12n-13)]/3
收起
不该是幂^,是*乘号吧