设数列an的前n项和Sn满足S(n+1)=4an+2,a1=1,(1)若bn=a(n+1)-2an,求bn表达式
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设数列an的前n项和Sn满足S(n+1)=4an+2,a1=1,(1)若bn=a(n+1)-2an,求bn表达式
(2)若cn=1/(a(n+1)-an),求数列cn所有项的和.
(2)若cn=1/(a(n+1)-an),求数列cn所有项的和.
(1)a[2]=5
s[n+2]-s[n+1]得
a[n+2]=4a[n+1]-4a[n]
a[n+2]-2a[n+1]=2a[n+1]-4a[n]
b[n+1]=2b[n]
b[1]=3
b[n]=3^2(n-1)
(2)设d[n]=a[n]/2^n
d[n+1]-d[n]=3/4
d[1]=1/2
d[n]=-1/4+3/4n
a[n]=(-1/4+3/4n)2^n
c[n]=1/(2^n(5/4+3/4n))
s[n]为c[n]的和
用错位相消法可解得s[n]
s[n+2]-s[n+1]得
a[n+2]=4a[n+1]-4a[n]
a[n+2]-2a[n+1]=2a[n+1]-4a[n]
b[n+1]=2b[n]
b[1]=3
b[n]=3^2(n-1)
(2)设d[n]=a[n]/2^n
d[n+1]-d[n]=3/4
d[1]=1/2
d[n]=-1/4+3/4n
a[n]=(-1/4+3/4n)2^n
c[n]=1/(2^n(5/4+3/4n))
s[n]为c[n]的和
用错位相消法可解得s[n]
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