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
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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
(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
![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](/uploads/image/z/8401967-71-7.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2Ca1%3D2%2Ca2%3D4%2C%E5%BD%93n%E2%89%A53%E6%97%B6%2CSn%2BS%28n-2%29%3D2%5BS%28n-1%29%5D%2B2)
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
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
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