Question

数学归纳法

这题怎么解啊 我算到K+1后就不会算了 哎

Answer 1

n = 1, 左边 = 1*3 = 3
右边 = [(2*1-1)(2*1 + 1)(2*1 +3) +3]/6 = (1*3*5 + 3)/6 = 18/6 = 3
左边=右边
设等式在n=k (k > 1)时成立, 1*3 + 3*5 + ... + (2k-1)(2k+1) = [(2k-1)(2k+1)(2k+3) + 3]/6
n=k+1时
1*3 + 3*5 + ... + (2k-1)(2k+1) + (2k+1)(2k+3)
= [(2k-1)(2k+1)(2k+3) + 3]/6 + (2k+1)(2k+3)
= [(2k-1)(2k+1)(2k+3) + 3 + 6(2k+1)(2k+3)]/6
= [(2k+1)(2k+3) (2k-1 + 6) + 3]/6
= [(2k+1)(2k+3) (2k +5) + 3]/6
即在n=k+1时也成立，于是对所有正整数都成立。