Question

一道数学题，急！！！！！！！！

1/(1*3)+1/(3*5)+1/(5*7)+~~~+n=？

Answer 1

裂项相消法做。

Answer 2

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

=1/2[1-1/(2n+1)]
=1/2[1-1/(2n+1)]
=n/(2n+1)

Answer 3

1/(1*3)+1/(3*5)+1/(5*7)+~~~+1/（n-1）n
=1/2（1-1/3+1/3-1/5+1/5-1/7+……-1/(n-1)+1/(n-1)-1/n）
=1/2(1-1/n)
=(n-1)/2n

Answer 4

题最后一项应该是1/（2n-1）(2n+1)
原式=(1-1/3)+(1/3-1/5)+(1/5-1/7)+……+[1/(2n-1)-1/(2n+1)]
=1-1/(2n+1)
=2n/(2n+1)

Answer 5

1/(1*3)+1/(3*5)+1/(5*7)+...+ 1/(n-2)*n
=0.5* (1/1 - 1/3 + 1/3 -1/5 +1/5 -1/7 +...-1/n)
=0.5*(1-1/n)

Answer 6

1/(1*3)+1/(3*5)+1/(5*7)+……+1/(2n-1)(2n+1)
=(1/2)*(1-1/3)+(1/2)*(1/3-1/5)+……+（1/2)*[1/(2n+1) -1/(2N-1)]
=(1/2)*[1-1/(2n-1)]
=1/2-1/(4n-2)