2019-2020学年北师大版选修2-2课时分层作业6 数学归纳法 作业(1)
2019-2020学年北师大版选修2-2课时分层作业6 数学归纳法 作业(1)第3页

  12+22+...+k2+(k+1)2=(k"(" k+1")(" 2k+1")" )/6+(k+1)2=(k"(" k+1")(" 2k+1")" +6"(" k+1")" ^2)/6

  =("(" k+1")(" 2k^2+7k+6")" )/6=("(" k+1")(" k+2")(" 2k+3")" )/6

  =("(" k+1")[(" k+1")" +1"][" 2"(" k+1")" +1"]" )/6,

  即当n=k+1时等式也成立.

  由(1)和(2),可知等式对任何n∈N+都成立.

7.已知正数数列{an}(n∈N+)的前n项和为Sn,且2Sn=an+1/a_n ,请用数学归纳法证明an=√n-√(n"-" 1).

证明(1)当n=1时,∵a1=S1=1/2 (a_1+1/a_1 ),

  ∴a_1^2=1(an>0).

  ∴a1=1.又√1-√0=1,∴当n=1时,结论成立.

  (2)假设当n=k(k≥1,k∈N+)时,结论成立,

  即ak=√k-√(k"-" 1),

  则当n=k+1时,

  ak+1=Sk+1-Sk=1/2 (a_(k+1)+1/a_(k+1) )-1/2 (a_k+1/a_k )

  =1/2 (a_(k+1)+1/a_(k+1) )-1/2 (√k "-" √(k"-" 1)+1/(√k "-" √(k"-" 1)))

  =1/2 (a_(k+1)+1/a_(k+1) )-√k.

  ∴a_(k+1)^2+2√kak+1-1=0,

  解得ak+1=√(k+1)-√k(an>0).

  ∴当n=k+1时,结论成立.

  由(1)和(2),可知对n∈N+都有an=√n-√(n"-" 1).

8.导学号88184008用数学归纳法证明:当n∈N+时,1+22+33+...+nn<(n+1)n.

证明(1)当n=1时,左边=1,右边=2,1<2,不等式成立.

  (2)假设当n=k(k≥1,k∈N+)时不等式成立,即1+22+33+...+kk<(k+1)k.

  则当n=k+1时,1+22+33+...+kk+(k+1)k+1<(k+1)k+(k+1)k+1=(k+1)k(k+2)<(k+2)k+1=[(k+1)+1]k+1.

  所以当n=k+1时不等式也成立.

  根据(1)和(2),可知不等式对任意n∈N+都成立.

B组