InterviewSolution
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InterviewSolution
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The sum of the series `1 + (1)/(1!) ((1)/(4)) + (1.3)/(2!) ((1)/(4))^(2) + (1.3.5)/(3!) ((1)/(4))^(3)+ ... ` to `infty`, isA. `sqrt(2)`B. 2C. `(1)/(sqrt(2))`D. none of these |
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Answer» Correct Answer - a Comparing the given series with the series `(1 + x)^(n) = 1 + (x)/(1!) + (n(n-1))/(2!)x^(2) + …., ` we get `nx = (1)/(4) and (n(n-1))/(2!) x^(2) = (1.3)/(2!)((1)/(4))^(2)` `rArr nx = (1)/(4) and n^(2) x^(2) - (nx) x = (3)/(16)` `rArr (1)/(16) - (x)/(4) = (3)/(16) rArr (x)/(4) = - (1)/(8) rArr x = - (1)/(2)` ` therefore nx = (1)/(4) rArr n = - (1)/(2)` Hence, sum of the series = `(1 - (1)/(2))^(-1//2) = sqrt(2)` |
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