设数列{an}满足：a（n+1）an=2a（n+1）-2（n=1,2,……）

Question

a2009=√2，则此数列的前2009项的和为______

Answer 1

a(n+1)a(n) = 2a(n+1) - 2,
特征方程为，r^2 = 2r - 2, 0 = r^2 - 2r + 2 = (r-1)^2 + 1,无实数根，{a(n)}为周期数列。

2 = 2a(n+1)-a(n+1)a(n) = a(n+1)[2 - a(n)],
a(n)不为2，
a(n+1) = 2/[2-a(n)].

记a(1)=a,则，
a(2) = 2/[2-a(1)] = 2/(2-a).
a(3) = 2/[2-a(2)] = 2/[2 - 2/(2-a)] = 2(2-a)/[4-2a-2] = 2(2-a)/(2-2a) = (2-a)/(1-a). [a不为1]
a(4) = 2/[2-a(3)] = 2/[2 - (2-a)/(1-a)] = 2(1-a)/[2-2a-2+a] = 2(1-a)/[-a] = 2(a-1)/a. [a不为0]
a(5) = 2/[2-a(4)] = 2/[2 - 2(a-1)/a] = 2a/[2a-2a+2] = a = a(1),
a(6) = 2/[2-a(5)] = 2/[2-a(1)] = a(2),
a(7) = 2/[2-a(6)] = 2/[2-a(2)] = a(3),
a(8) = 2/[2-a(7)] = 2/[2-a(3)] = a(4),
...
a(4n-3) = a(1) = a,
a(4n-2) = a(2) = 2/(2-a),
a(4n-3) = a(3) = (2-a)/(1-a),
a(4n) = a(4) = 2(a-1)/a.

2^(1/2) = a(2009) = a(4*503 - 3) = (2-a)/(1-a),
2^(1/2) - a*2^(1/2) = 2 - a,
2^(1/2) - 2 = [2^(1/2) - 1]*a,
a = -2^(1/2),

a(4n-3) = -2^(1/2),
a(4n-2) = 2/(2-a) = 2/[2+2^(1/2) ]= 2[2-2^(1/2)]/[4-2] = 2-2^(1/2),
a(4n-3) = (2-a)/(1-a) = [2+2^(1/2)]/[1+2^(1/2)] = 2^(1/2),
a(4n) = 2(a-1)/a = 2[-2^(1/2)-1]/[-2^(1/2)] = 2^(1/2)[2^(1/2)+1] = 2 + 2^(1/2).

a(4n-3) + a(4n-2) + a(4n-3) + a(4n) = -2^(1/2) + 2 - 2^(1/2) + 2^(1/2) + 2 + 2^(1/2) = 4,

a(1)+a(2)+...+a(2008)+a(2009)
= [a(4*1-3)+a(4*1-2)+a(4*1-1)+a(4*1)] + [a(4*2-3) + a(4*2-2) + a(4*2-1) + a(4*2)] + ... + [a(4*502-3) + a(4*502-2) + a(4*502-1) + a(4*502)] + a(2009)
= 4*502 + a(2009)
= 2008 + 2^(1/2)

Answer 2

1，a(n+1)*an-2a(n+1)+1=0
a(n+1)*(2-an)=1
所以a(n+1)=1/(2-an)
那么1-a(n+1)=1-1/(2-an)=(1-an)/(2-an)
所以1/[1-a(n+1)]=(2-an)/(1-an)=1+1/(1-an)
于是1/[1-a(n+1)]-1/(1-an)=1，为常数
而1/(1-a1)=1/(1-0)=1
所以数列{1/(1-an)}时以1为首项、1为公差的等差数列
2，1/(1-an)=1+(n-1)×1=n，所以an=(n-1)/n (n∈N+)
3，bn={1-√[n/(n+1)]}/√n=1/√n-1/√(n+1)
所以Sn=1-1/√2+1/√2-1/√3+1/√3-1/√4+……+1/√n-1/√(n+1)
=1-1/√(n+1)
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