Two Ways to Find a FormulaDate: 04/14/98 at 17:24:19 From: Mon Subject: Calculus, differentiating, geometric series For any real number x, such that x does not equal 1, I am trying to prove that: n x-(n+1)x^(n+1) + nx^(n+2) Sigma rx^r = ------------------------- r=1 (1-x)^2 by using calculus, differentiating (with respect to x), and the formula for the sum of the geometric series: n Sigma x^r r=0 Date: 04/15/98 at 08:02:27 From: Doctor Anthony Subject: Re: Calculus, differentiating, geometric series The series required is: x + 2x^2 + 3x^3 + 4x^4 + ... + nx^n You can sum this in two ways. The non-calculus way is as follows: S = x + 2x^2 + 3x^3 + ... + nx^n xS = x^2 + 2x^3 + ... + (n-1)x^n + nx^(n+1) ----------------------------------------------------- subtracting (1-x)S = x + x^2 + x^3 + ... + x^n - nx^(n+1) (1-x)S = x(1-x^n)/(1-x) - nx^(n+1) x(1-x^n) nx^(n+1) S = -------- - ---------- (1-x)^2 (1-x) x(1-x^n) - n(1-x)x^(n+1) S = ------------------------ (1-x)^2 x - x^(n+1) - nx^(n+1) + nx^(n+2) S = --------------------------------- (1-x)^2 x - (n+1)x^(n+1) + nx^(n+2) S = ---------------------------- (1-x)^2 CALCULUS METHOD Consider the series: 1 + x + x^2 + ......+ x^n differentiating: 1 + 2x + 3x^2 + 4x^3 + .. + nx^(n-1) multiply by x: x + 2x^2 + 3x^3 + ........+ nx^n The sum of the series 1 + x + x^2 + .... + x^n is given by: 1-x^(n+1) --------- 1-x and differentiating this we get: (1-x)(-(n+1)x^n) - (1-x^(n+1))(-1) ---------------------------------- (1-x)^2 -(n+1)x^n + (n+1)x^(n+1) + 1 - x^(n+1) = -------------------------------------- (1-x)^2 1 - (n+1)x^n + nx^(n+1) = ----------------------- (1-x)^2 and finally to get the sum we require we must multiply this result by x. The required sum is: x - (n+1)x^(n+1) + nx^(n+2) --------------------------- (1-x)^2 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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