是否存在常数a,b使等式1^2⼀1*3+2^2⼀3*+.......+n^2⼀(2n-1)(2n+1)=an^2+n⼀bn+2对一切正实数都成立

请高手解答 。。。。 要是不会别瞎回答 这是作业 跪求
2025-01-05 17:04:25
推荐回答(1个)
回答1:

n^2/(2n-1)(2n+1)=(n^2-1/4+1/4)/(4n^2-1)=1/4+1/4(1//(2n-1)(2n+1))
=1/4+1/8(1/(2n-1)-1/(2n+1))
故原式
=1/4*n+1/8(1-1/3+1/3-1/5+.....+(1/(2n-1)-1/(2n+1))
=n/4+1/8(1-1/(2n+1))
=n/4+n/(8n+4)
=(n^2+n)/(4n+2)=(an^2+n)/(bn+2)
显然a=1 b=4 以后把题写清楚
1^2/1*3+2^2/3*+.......+n^2/(2n-1)(2n+1)=(an^2+n)/(bn+2)加上括号,否则容易引起歧义