Рекуррентные вычисления

Private Sub CommandButton1_Click()
Dim n As Long, i As Long
Dim s As Double, a As Double, x As Double
n = Val(TextBox1.Text)
x = Val(TextBox2.Text)
a = 1
s = 0
For i = 0 To n
s = s + a
a = -a * x ^ 2 * (3 ^ (2 * i + 2) + 3) / (2 * i + 1) * (2 * i + 2) * (3 ^ (2 * i) + 3)
Next i
TextBox3.Text = Str(s)
End Sub
Private Sub CommandButton2_Click()
Dim i As Integer
Dim s As Single, a As Single
Dim st As Single, e As Single
e = Val(TextBox5.Text)
x = Val(TextBox4.Text)
a = 1
s = 0
i = 0
st = (Cos(x)) ^ 3
While Abs(s - st) >= e
s = s + a
a = a = -a * x ^ 2 * (3 ^ (2 * i + 2) + 3) / (2 * i + 1) * (2 * i + 2) * (3 ^ (2 * i) + 3)
i = i + 1
Wend
TextBox6.Text = Str(i - 1)
TextBox10.Text = Str(s)
End Sub
 
Private Sub CommandButton3_Click()
Dim i As Integer, e As Single
Dim s As Single, a As Single, x As Single
e = Val(TextBox8.Text)
x = Val(TextBox7.Text)
a = 1
s = 0
i = 0
While Abs(a) >= e
s = s + a
a = a = -a * x ^ 2 * (3 ^ (2 * i + 2) + 3) / (2 * i + 1) * (2 * i + 2) * (3 ^ (2 * i) + 3)
i = i + 1
Wend
TextBox9.Text = Str(i - 1)
TextBox11.Text = Str(s)
End Sub

Function Cos3(ByVal x As Double, Optional ByVal e As Double = 0.000001) As Double
Dim i As Long, triN As Double, xN_znam As Double, a As Double
  e = e * 4   'сумма ряда потом делится на 4, так что допуск можно увеличить в 4 раза
  x = -x * x  '-x^2 для умножения в цикле
  triN = 1    'степень тройки
  xN_znam = 1 'степень х, деленная на знаменатель
  Do
    a = (triN + 3) * xN_znam 'очередной член ряда
    Cos3 = Cos3 + a
    If Abs(a) < e Then Exit Do
    triN = triN * 9
    i = i + 2
    xN_znam = xN_znam * x / (i - 1) / i
  Loop
  Cos3 = Cos3 / 4
End Function

Sub test()
Dim x#, f#, c#, d$
  d = "0.00000000"
  x = 0.1
  Debug.Print "x", "ряд", "Cos(x) ^ 3", "погрешность"; vbLf; String(50, "-")
  Do While x < 11
    f = Cos3(x): c = Cos(x) ^ 3
    Debug.Print Format(x, d), Format(f, d), Format(c, d), Format(Abs(f - c), "0.0E-00")
    x = x * 1.25
  Loop
End Sub