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Formula : An = A1r^n-1
Examples : 1, 2, 4, 8, 16 r = 2 1, 3, 9, 27, 81 r = 3 1, 5, 25, 125, 625 r = 5
Formula : Sn = A1 (1-r^n) (-------) ( 1-r ) or Sn = A1 (r^n-1) (-------) ( r-1 ) Formula : (Infinite) Sn = A1/r-1 en
Example : The Cantor Set. What is the length of the Cantor Set? The easiest way to answer this question is to compute the length of the “middle third” intervals which are removed at each stage of creating the Cantor Set. The first middle third has length 1/3. The second middle third has length 1/9, but since we remove the middle thirds of two subintervals, the total length removed at the second stage is 2/9. At the third stage we remove four middle thirds, each of length 1/27 , so the total length removed is 4/27 . Since these middle thirds do not overlap, we see that the total length of the intervals removed is: 1/3 + 2/9 +4/27 + 8/81 +....... Once again, this is a geometric series, with first term a = 1/3 and common ratio r = 2/3. By the formula, the total length L of all these middle thirds is L= 1/3 + 2/9 + 4/27 + 8/81 + .... = (1/3)/1-(2/3) = (1/3)/(1/3) = 1 If we begin with the unit interval [0, 1] and remove intervals whose total length is 1, the points that remain—in this case, the Cantor set—must have length 1 − 1, or 0. The answer to Question 1 at the end of the Cantor Set Section demonstrates that there are as many points in the Cantor set as there points in the interval [0, 1]. It boggles the mind to think that they all fit in a space whose total length is zero.
Formula: ∞ ∑ = 1/n n =1
Example: 1 + 1/2 + 1/3 + 1/4 + 1/5 + ...
Formula: an = 1/n
Example: 1, 1/2, 1/3, 1/4, 1/5
Formula : Sn = n/2 (A1+An)
Examples : 1, 2, 3, 4, 5 S5 = 15 16, 33, 50, 67, 84 S5 = 250 1, -3, -7, -11, -15 S5 = -35
Formula : An = A1 + (n-1) d
Examples : 1, 2, 3, 4, 5 d = 1 16, 33, 50, 67, 84 d = 17 1, -3, -7, -11, -15 d = -4