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