Question

已知数列{an}的前n项和为Sn,a1=2,a2=4,当n≥3时，Sn+S(n-2)=2[S(n-1)]+2

(1)求证数列{an}是等差数列。
(2)设数列{bn}对任意的n∈N,，均有an=b1*S1+b2*S2+……+bn*Sn成立，求b1+b2+……+b2011的值

Answer 1

Sn+S(n-2)=2[S(n-1)]+2
Sn-[S(n-1)] = [S(n-1)] - S(n-2) +2
an = an-1 +2
an - an-1 = 2
n>=3
a1=2,a2=4,
所以 {an}是等差数列。
数列{bn}对任意的n∈N,，均有an=b1*S1+b2*S2+……+bn*Sn成立
所以
an=b1*S1+b2*S2+……+bn*Sn (1)
an+1 = b1*S1+b2*S2+……+bn*Sn + b(n+1)*S(n+1) (2)
(2) - (1)
b(n+1)*S(n+1) = 2
b.n*Sn = 2
bn = 2/Sn
= 2/[ n( a1 + an) /2 ]
=4 / [ n( 2 + 2(2+2(n-1)) ) ]
=2/n*(n+1)
b1+b2+……+b2011
= 2 [ 1/(1x2 ) + 1/(2x3 ) +…… + 1/(2011x2012 ) ]
=2[ 1-1/2 + 1/2 - 1/3 +1/3 -1/4 + …… +1/2011 - 1/2012 ]
= 2[ 1- 1/2012 ]
=2011/1006

Answer 2

2) - (1)
b(n+1)*S(n+1) = 2
b.n*Sn = 2
bn = 2/Sn
= 2/[ n( a1 + an) /2 ]
=4 / [ n( 2 + 2(2+2(n-1)) ) ]
=2/n*(n+1)
b1+b2+……+b2011
= 2 [ 1/(1x2 ) + 1/(2x3 ) +…… + 1/(2011x2012 ) ]
=2[ 1-1/2 + 1/2 - 1/3 +1/3 -1/4 + …… +1/2011 - 1/2012 ]
= 2[ 1- 1/2012 ]
=2011/1006