数列如图
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数列如图
数列如图
数列如图
a1=2,
a(n+1) = 2^(n+1).an/( n+ (1/2)an+2^n)
(1)
a(n+1) = 2^(n+1).an/( n+ (1/2)an+2^n)
1/a(n+1) = ( n+ (1/2)an+2^n)/[2^(n+1).an]
2^(n+1)/a(n+1) = [ 2^n/an + n + 1/2 ]
2^(n+1)/a(n+1) - 2^n/an = n+(1/2)
2^n/an - 2^(n-1)/a(n-1) = n-(1/2)
2^n/an - 2/a1 = (n+2)(n-1)/2 - (n-1)/2
= (n^2-1)/2
2^n/an = (n^2+1)/2
an = 2^(n+1)/(n^2+1)
bn = 2^n/an
= (n^2+1)/2
(2)
cn = 1/[n(n+1)a(n+1)]
=(1/[n(n+1)]) .( (n+1)^2+1)/ 2^(n+2)
= (1/[n(n+1)]) (n^2+2n+2)/2^(n+2)
= [1/2^(n+2)] [ 1 +(n+2)/(n(n+1) )]
consider
(n+2)/(n(n+1) ) = a/n + b/(n+1)
n+2 = a(n+1) +bn
put n=0,=> a= 2
put n= -1 => b = -1
cn = [1/2^(n+2)] [ 1 +2/n - 1/(n+1)]
= 1/2^(n+2) + (1/2^(n+2))( 2/n -1/(n+1) )
Sn = c1+ c2 +..+cn
= (1/4)( 1- (1/2)^n ) + summation(i:1->n ) (1/2^(i+2))( 2/i -1/(i+1) )
=(1/4)( 1- (1/2)^n ) + [ ( 1/4 - 1/16) + (1/16 -1/48) + ..+(1/2^(n+2))( 2/n -1/(n+1) ) ]
=(1/4)( 1- (1/2)^n ) + 1/4 - 1/[(n+1).2^(n+2)]