Question

若1/1x3+1/3x5+1/5x7+……+1/(2n-1)(2n+1)等于17/35，求n的值

Answer 1

1/1x3+1/3x5+1/5x7+……+1/(2n-1)(2n+1)
＝1/2((1-1/3)+(1/3-1/5)+....+(1/(2n-1)-1/(2n+1)) 中间可消去
=1/2(1-1/(2n+1))
又因为＝17/35
所以n=17

Answer 2

1/(2n-1)(2n+1)= 0.5[1/(2n-1)-1/(2n+1)]
故1/1x3+1/3x5+1/5x7+……+1/(2n-1)(2n+1) = 0.5[1-1/3+1/3-1/5+1/5-1/7+...1/(2n-1)-1/(2n+1)]
=0.5(1-1/(2n+1)]=n/(2n+1)
n/(2n+1) =17/35
得n＝17

Answer 3

用裂项相消法1/（1x3） 1/（3x5） 1/（5x7） …… 1/(2n-1)(2n 1)＝1/2*[1-1/3 1/3-1/5 … 1/(2n-1)-1/(2n 1)]=1/2*[1-1/(2n 1)]=n/(2n 1)=17/35解得n=17

Answer 4

1/2[1-1/3+1/3-1/5+1/5-1/7+……+1/（2n-1)-1/(2n+1)]=17/35
1-1/(2n+1)=34/35
2n+1=35
n=17

Answer 5

1/1*3+1/3*5+1/5*7+...1/(2n-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)]
=n/(2n+1)
∴n/(2n+1）=17/35 ∴n=17