求和裂项相消(2눀⼀1*3)+(4눀⼀3*5)+(6눀⼀5*7)+....+[﹙2n﹚눀⼀(2n-1)*(2n+1)]

2024-11-29 04:51:04
推荐回答(2个)
回答(1):

(2²/1*3)+(4²/3*5)+(6²/5*7)+....+[﹙2n﹚²/(2n-1)*(2n+1)]

=n+1/1*3+1/3*5)+1/5*7+....+1/(2n-1)*(2n+1)
=n+[1-1/3+1/3-1/5+1/5-1/7+……+1/(2n-1)-1/(2n+1)]÷2
=n+[1-1/(2n+1)]÷2
=n+2n/(2n+1)÷2
=n+n/(2n+1)

回答(2):

(2²/1*3)+(4²/3*5)+(6²/5*7)+....+[﹙2n﹚²/(2n-1)*(2n+1)]
=n+1/1*3+1/3*5)+1/5*7+....+1/(2n-1)*(2n+1)
=n+[1-1/3+1/3-1/5+1/5-1/7+……+1/(2n-1)-1/(2n+1)]÷2
=n+[1-1/(2n+1)]÷2
=n+2n/(2n+1)÷2
=n+n/(2n+1)