高中数学 数列求和

Cn=(n+1)/(2^n)
Cn=n/(2^n)+1/2^n
Sn=Sn1+Sn2
Sn1-2Sn1=1/2+2/4+3/8+...-[1+2/2+3/4+..+n/2^(n-1)]
=-[1+1/2+1/4+..+1/2^(n-1)]+n/(2^n)
=-2[1-1/(2^n)]+n/(2^n)=-Sn1
Sn1=2[1-1/(2^n)]-n/(2^n)
Sn2=1-1/(2^n)
Sn=3[1-1/(2^n)]-n/(2^n)

用错位相消法
Sn=2/2+3/(2^2)+4/(2^3)+5/(2^4)+.....+(n-1)/[2^(n-2)]+n/[2^(n-1)]+(n+1)/(2^n) A
(1/2)Sn=2/(2^2)+3/(2^3)+4/(2^4)+.....+n/[2^(n-2)]+(n+1)/[2^(n-1)]+(n+2)/(2^n)+(n+2)/[2^(n+1)] B
用B式减A式
-（1/2）Sn=-2/2+{1/2^2+1/2^3+....+1/2^n}+(n+2)/[2^(n+1)]
1/2^2+1/2^3+....+1/2^n 可用等比数列求和公式Sn=[a1(1-q^n)]/(1-q)来求
其中a1=1/2^2 q=1/2 n=n-1(因为从1/2^2到1/2^n共n-1项)
1/2^2+1/2^3+....+1/2^n =1/2+1/2^n
故-（1/2）Sn==-2/2+1/2+1/2^n+(n+2)/[2^(n+1)]=-1/2+1/2^n+(n+2)/[2^(n+1)]
Sn=1/2^2-1/2^(n+1)-(n+2)/[2^(n+2)]