設一數列首項為6，第n+1項=1/(第n項+1)，n為正整數，則第2011項減第2010項等於多少？

令此數列為 An , 由題意知:
A1 = 6 且 An+1 = 1/( An + 1 )

A2 = 1/(6+1) = 1/7
A3 = 1/( 1/7 + 1 ) = 7/8
A4 = 1/( 7/8 + 1 ) = 8/15
A5 = 1/( 8/15 + 1 ) = 15/23

可歸納得 :
An+1 = Pn / Pn+1
其中 Pn 滿足: Pn+2 = Pn + Pn+1
也就是, 此數列為有理數, 分子為前一項的分母, 分母形成遞迴數列.
證明請參考底下的註解1

Pn+2 = Pn + Pn+1
Pn+2 - Pn+1 - Pn = 0 , 此遞迴關係的特徵方程式為:
x^2 - x - 1 = 0
x = ( 1 ± √5 )/2 , 所以:
Pn = c1*α^n + c2*β^n , 其中 α = ( 1 + √5 )/2 , β = ( 1 - √5 )/2

A1 = 6/1 , 所以 P1 = A1的分母 = 1
A2 = 1/7 , 所以 P2 = 7
代入初始值 P1 = 1 與 P2 = 7 可解得:
c1 = 3 - 0.4√5 , c2 = 3 + 0.4√5 ..... 註解2

An+1 - An
= Pn/Pn+1 - Pn-1/Pn
= [ (Pn)^2 - Pn-1*Pn+1 ] / ( Pn+1 * Pn )
= (-1)^n * 41 / ( Pn+1 * Pn ) ..... 註解3

當 n = 2010
A2011 - A2010
= (-1)^2010 * 41 / ( P2011 * P2010 )
= 41 / ( P2011 * P2010 )

Ans:
第2011項減第2010項 = 41 / ( P2011 * P2010 )
其中 :
Pn = c1*α^n + c2*β^n
c1 = 3 - 0.4√5 , c2 = 3 + 0.4√5
α = ( 1 + √5 )/2 , β = ( 1 - √5 )/2
應注意的是分母 ( P2011 * P2010 ) 為整數, 原因請參考註解5

---------------------------------------------------------------------------

註解1
claim : An+1 = Pn / Pn+1
其中 Pn 滿足: Pn+2 = Pn + Pn+1
pf :
令 An = Qn / Pn
Qn+1 / Pn+1 = An+1 = 1/( An + 1 ) = 1/[ (Qn/Pn) + 1 ] = Pn / ( Pn + Qn )
所以: Qn+1 = Pn 且 Pn+1 = Pn + Qn
An+1 = Qn+1 / Pn+1 = Pn / Pn+1
Pn+2 = Pn+1 + Qn+1 = Pn+1 + Pn

註解2
claim : c1 = 3 - 0.4√5 , c2 = 3 + 0.4√5
pf :
P1 = c1*α + c2*β = c1*[ ( 1 + √5 )/2 ] + c2*[ ( 1 - √5 )/2 ] = 1
c1*( 1 + √5 ) + c2*( 1 - √5 ) = 2
( c1 + c2 ) + √5( c1 - c2 ) = 2 ..... (1式)

P2 = c1*α^2 + c2*β^2 = c1*[ ( 1 + √5 )/2 ]^2 + c2*[ ( 1 - √5 )/2 ]^2 = 7
c1*( 6 + 2√5 ) + c2*( 6 - 2√5 ) = 28
6( c1 + c2 ) + 2√5( c1 - c2 ) = 28
3( c1 + c2 ) + √5( c1 - c2 ) = 14 ..... (2式)

(2式) - (1式) 得:
2( c1 + c2 ) = 12 , 所以 c1 + c2 = 6
(1式)*3 - (2式) 得:
2√5( c1 - c2 ) = - 8 , 所以 c1 - c2 = - 8/(2√5) = - 4/√5 = - 4√5 / 5 = - 0.8√5
c1 = ( 6 - 0.8√5 )/2 = 3 - 0.4√5
c2 = ( 6 + 0.8√5 )/2 = 3 + 0.4√5

註解3
claim : (Pn)^2 - Pn-1*Pn+1 = (-1)^n * 41
pf :
(Pn)^2 - Pn-1*Pn+1
= ( c1*α^n + c2*β^n )^2 - [ c1*α^(n-1) + c2*β^(n-1) ]*[ c1*α^(n+1) + c2*β^(n+1) ]
= (c1)^2*α^(2n) + (c2)^2*β^(2n) + 2*c1*c2*α^n*β^n
- (c1)^2*α^(2n) - (c2)^2*β^(2n) - c1*c2*α^(n-1)*β^(n-1)*( α^2 + β^2 )
= - c1*c2*α^(n-1)*β^(n-1)*( α^2 + β^2 - 2αβ )
= - c1*c2 * (αβ)^(n-1) * ( α - β )^2
= - (41/5) * (-1)^(n-1) * 5 ..... 註解4
= (-1)^n * 41

註解4
c1*c2 = ( 3 - 0.4√5 )( 3 + 0.4√5 ) = 9 - 0.16*5 = 8.2 = 41/5
αβ = [ ( 1 + √5 )/2 ] * [ ( 1 - √5 )/2 ] = ( 1 - 5 )/4 = - 1
α - β = [ ( 1 + √5 )/2 ] - [ ( 1 - √5 )/2 ] = √5

註解5
因為初始值 P1 = 1 與 P2 = 7 皆為整數,
又 Pn+2 = Pn + Pn+1
所以對所有自然數n , Pn皆為整數
所以 ( P2011 * P2010 ) 為整數