已知数列an+1=2an/an+1,且a1=1/2,则a5=
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a(n+1) = 2an/(an+1)
1/a(n+1) = (an+1)/(2an)
= 1/2 + 1/(2an)
1/a(n+1) -1 = (1/2) ( 1/an -1 )
{1/an -1 } 是等比数列,q=1/2
1/an- 1 = (1/2)^(n-1) .( 1/a1 - 1)
= (1/2)^(n-1)
an = 1/[ 1+ (1/2)^(n-1)]
a5 = 1/[ 1+ (1/2)^4]
= 16/17
1/a(n+1) = (an+1)/(2an)
= 1/2 + 1/(2an)
1/a(n+1) -1 = (1/2) ( 1/an -1 )
{1/an -1 } 是等比数列,q=1/2
1/an- 1 = (1/2)^(n-1) .( 1/a1 - 1)
= (1/2)^(n-1)
an = 1/[ 1+ (1/2)^(n-1)]
a5 = 1/[ 1+ (1/2)^4]
= 16/17
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