Find the sum of the series 1+(1/3)+(1.3/3.6)+(1.3.5/3.6.9)+.......... is
MERIN VARGHESE , 5 Years ago
Grade
Algebra> Find the sum of the series 1+(1/3)+(1.3/3...
1 Answers
Arun
Last Activity: 5 Years ago
This is simply a particular case of the binomial series (follow the link below to read an article):
(1 + x)^n = ∑[k=0 .. ∞] n(n-1)(n-2) . . (n-k+1) x^k / k!,
having n = -1/2 and x = -2/3:
(1 - 2/3)^(-1/2) = 1 + (-1/2) * (-2/3) + (-1/2)(-3/2) * (-2/3)^2/2! +
+ (-1/2)(-3/2)(-5/2) * (-2/3)^3/3! + (-1/2)(-3/2)(-5/2)(-7/2) * (-2/3)^4/4! + . . =
= 1 + 1/3 + (1*3/2^2)(2^2/3^2)/2! + (1*3*5/2^3)(2^3/3^3)/3! + (1*3*5*7/2^4)(2^4/3^4)/4! + . . . =
/cancel the degrees of 2 now/
= 1 + 1/3 + 1*3/(3^2*2!) + 1*3*5/(3^3*3!) + 1*3*5*7/(3^4*4!) + . . =
= 1 + 1/3 + 1*3/(3*6) + 1*3*5/(3*6*9) + 1*3*5*7/(3*6*9*12) + . . . = (1/3)^(-1/2) = √3
(1 + x)^n = ∑[k=0 .. ∞] n(n-1)(n-2) . . (n-k+1) x^k / k!,
having n = -1/2 and x = -2/3:
(1 - 2/3)^(-1/2) = 1 + (-1/2) * (-2/3) + (-1/2)(-3/2) * (-2/3)^2/2! +
+ (-1/2)(-3/2)(-5/2) * (-2/3)^3/3! + (-1/2)(-3/2)(-5/2)(-7/2) * (-2/3)^4/4! + . . =
= 1 + 1/3 + (1*3/2^2)(2^2/3^2)/2! + (1*3*5/2^3)(2^3/3^3)/3! + (1*3*5*7/2^4)(2^4/3^4)/4! + . . . =
/cancel the degrees of 2 now/
= 1 + 1/3 + 1*3/(3^2*2!) + 1*3*5/(3^3*3!) + 1*3*5*7/(3^4*4!) + . . =
= 1 + 1/3 + 1*3/(3*6) + 1*3*5/(3*6*9) + 1*3*5*7/(3*6*9*12) + . . . = (1/3)^(-1/2) = √3
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