已在另一贴给你回答了
http://expert.csdn.net/Expert/topic/1470/1470061.xml?temp=2.262515E-02
http://expert.csdn.net/Expert/topic/1470/1470061.xml?temp=2.262515E-02
解决方案 »
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Next i
Dim SngSum As Single
Dim SngIndex As Single
For IntIndex = 1 To SngNumber
If (IntIndex Mod 2) = 0 Then
SngSum = SngSum - (IntIndex - 1) / IntIndex
Else
SngSum = SngSum + IntIndex
End If
Next
Dim fz As Double, _
fm As Double, _
sum As Double
If n <= 0 Then Exit Function
If n = 1 Then
sumx = 1
Else
sum = 1
fz = 1
fm = 2
For i = 1 To n - 1
If i Mod 2 > 0 Then
sum = sum - fz / fm
Else
sum = sum + fz / fm
End If
fz = 1 + i
fm = fz + i
Next
sumx = sum
End If
End Function
上面三位老兄,请你们仔细看清楚一点,好吗?这道题目应该是:
1-1/2+2/3-3/5+5/8-8/13……
它的规律是:前两位数的分子之和是后一位数的分子,而前两位数的分母之和是后一位数的分母。
而上面那位老兄做的却是:
1-1/2+2/3-3/5+4/7-5/9……
有何规律而言呢?恳请各位高手指教!多谢了!
fz1=1
fz2=1
fm1=1
fm2=2
for i=1 to 20
sum=fz1/fm1-fz2/fm2+sum
fz1=fz1+fz2
fm1=fm1+fm2
fz2=fz1+fz2
fm2=fm1+fm2
next
Dim t(1 To 30) As Long
Dim i As Integer
t(1) = 1
t(2) = 1
For i = 3 To 30
t(i) = t(i - 1) + t(i - 2)
Next i
For i = 1 To 30
Debug.Print t(i)
Next i
Dim sum As Double
sum = 1
For i = 1 To 20
sum = sum + ((-1) ^ i) * (t(i + 1) / t(i + 2))
Debug.Print "sum="; IIf(((-1) ^ i) > 0, "+", "-"); "("; t(i + 1); "/"; t(i + 2); ")"
Next i
Debug.Print sum
End Sub
------------------
1
1
2
3
5
8
13
21
34
55
89
144
233
377
610
987
1597
2584
4181
6765
10946
17711
28657
46368
75025
121393
196418
317811
514229
832040
sum=-( 1 / 2 )
sum=+( 2 / 3 )
sum=-( 3 / 5 )
sum=+( 5 / 8 )
sum=-( 8 / 13 )
sum=+( 13 / 21 )
sum=-( 21 / 34 )
sum=+( 34 / 55 )
sum=-( 55 / 89 )
sum=+( 89 / 144 )
sum=-( 144 / 233 )
sum=+( 233 / 377 )
sum=-( 377 / 610 )
sum=+( 610 / 987 )
sum=-( 987 / 1597 )
sum=+( 1597 / 2584 )
sum=-( 2584 / 4181 )
sum=+( 4181 / 6765 )
sum=-( 6765 / 10946 )
sum=+( 10946 / 17711 )
1.19595578513638