归纳法求证1^2⼀(1*3)+2^2⼀(3*5)+...+n^2⼀(2n-1)(2n+1)=n(n+1)⼀[2(2n+1)]

2024-11-27 01:44:29
推荐回答(1个)
回答1:

当n=1时,左边=1^2/(1*3)=1/3,右边=1*2/(2*3)=1/3
成立
假设当n=K时
1^2/(1*3)+2^2/(3*5)+...+k^2/(2k-1)(k+1)=k(k+1)/[2(2k+1)]
成立
当n=k+1时
左边=k(k+1)/[2(2k+1)] +(k+1)^2/(2k+1)(k+2)
=(k+1)(k+2)/[2(2k+3)]
=(k+1)[(k+1)+1]/{2[2(k+1)+1]}=右边
所以1^2/(1*3)+2^2/(3*5)+...+n^2/(2n-1)(2n+1)=n(n+1)/[2(2n+1)]