Question

1乘3分之一加3乘5分之一加5乘7分之一一直加到二n加1，二n减1分之一等于多少

求结果

Answer 1

1/1*3+1/3*5+……+1/(2n-1)(2n+1)
=[2/1*3+2/3*5+……+2/(2n-1)(2n+1)]/2
={(1/1-1/3)+(1/3-1/5)+(1/5-1/7)+……+[1/(2n-1)-1/(2n+1)]}/2
=[1-1/(2n+1)]/2
=n/(2n+1)
1/（1*3）+1/（3*5）+1/（5*7）+...+1/（2n-1）*(2n+1)
=1/2[1/1-1/3+1/3-1/5+1/5-1/7+...+1/(2n-1)-1/(2n+1)](相临2项加减相互抵消就剩下第一项和最后一项)
=1/2(1-1/(2n+1)]
=n/(2n+1)

Answer 2

楼主打错了吧，应该是（2N-1）*(2n+1)分之一吧
如果是这样结果是（N+2分之N）

Answer 3

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

Answer 4

1/（1*3）+1/（3*5）+1/（5*7）+...+1/（2n-1）*(2n+1)
=1/2[1/1-1/3+1/3-1/5+1/5-1/7+...+1/(2n-1)-1/(2n+1)](相临2项加减相互抵消就剩下第一项和最后一项)
=1/2(1-1/(2n+1)]
=n/(2n+1)

Answer 5

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