(1)依题意:an + 1 =" bn" + 1 + cn + 1 =" a" an + an + b an2,
则a2 =" a" a1 + a1 + b a12 ∴a + 1 + b =
①
则a3 =" a" a2 + a2 + b a22 ∴
②
解①②得a = 1,b = –
从而an + 1 =" 2an" –
an2 (n∈N*) ………………………5分
(2)证法(Ⅰ)由于an + 1 =" 2an" –
an2 = –
(an – 2)2 + 2≤2.
但an + 1≠2,否则可推得a 1=" a" 2= 2与a 1= 1,a2 = 1.5矛盾.故an + 1<2 于是an
<2
又an + 1– an= –
an2 + 2an – an = –
an (an – 2) >0,
所以an + 1>an 从而an<an + 1<2 …………………………………9分
证法(Ⅱ)由数学归纳法
(i)当n = 1时,a1 = 1,a2 = 1.5,显然a1<a2<2成立
(ii)假设n = k时, ak<ak + 1<2成立.
由于函数f (x) = –
x2 + 2x = –
(x – 2)2 + 2在[0,2]上为增函数,
则f (ak) <f (ak + 1) <f (2)即
ak (4 – ak) <
ak + 1(4 –ak + 1) <
×2×(4 – 2)
即 ak + 1<ak + 2<2成立. 综上可得n∈N*有an<an + 1<2 …………………………9分
(3)由an + 1 =" 2an" –
an2得2 (an + 1– 2) =" –" (an – 2)2 即(2 – an + 1) =
(2 – an)2
又由(2)an<an + 1<2可知2 – an + 1>0,2 – an>0
则lg (2 – an + 1) =" 2" lg (2 – an) – lg 2 ∴lg (2 – an +1) – lg2 =" 2[lg" (2 – an) – lg2]
即{lg (2 – an + 1) – lg2}为等比数列,公比为2,首项为lg (2 – a1) – lg 2 =" –lg" 2
故lg (2 – an) – lg 2 =" (–lg" 2)·2n – 1 ∴an =" 2" – 2
(n∈N*)为所求………13分