2018-2019学年人教B版选修2-2 2.3数学归纳法 作业
2018-2019学年人教B版选修2-2 2.3数学归纳法 作业第3页

  令n=1,2,得{■(a"(" b+1")" =1"," @a"(" 4b+1")" =3"," )┤解得{■(a=1/3 "," @b=2"." )┤

  下面用数学归纳法证明a=1/3,b=2时等式对一切n∈N+都成立.

  (1)当n=1时,已证.

  (2)假设当n=k(k∈N+)时等式成立,即

  12+22+32+...+k2+(k-1)2+...+22+12=1/3 k(2k2+1),

  则当n=k+1时,12+22+...+k2+(k+1)2+k2+(k-1)2+...+22+12

  =1/3 k(2k2+1)+(k+1)2+k2

  =1/3 k(2k2+3k+1)+(k+1)2

  =1/3 k(2k+1)(k+1)+(k+1)2

  =1/3(k+1)(2k2+4k+3)

  =1/3(k+1)[2(k+1)2+1].

  故当n=k+1时,等式也成立.

  由(1)和(2),知存在a=1/3,b=2,使等式对一切n∈N+都成立.

9已知在数列{an}中,a1=2,an+1=(√2-1)(an+2),n=1,2,3,....

(1)求{an}的通项公式;

(2)若在数列{bn}中,b1=2,bn+1=(3b_n+4)/(2b_n+3),n=1,2,3,....证明:√2

(1)解由题设an+1=(√2-1)(an+2)

  =(√2-1)(an-√2)+(√2-1)(2+√2)

  =(√2-1)(an-√2)+√2,

  所以an+1-√2=(√2-1)(an-√2).

  所以数列{an-√2}是首项为2-√2,公比为√2-1的等比数列,则an-√2=√2(√2-1)n,

  即an的通项公式为an=√2[(√2-1)n+1],n=1,2,3,....

(2)证明用数学归纳法证明.

①当n=1时,因为√2<2,b1=a1=2,