InterviewSolution
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InterviewSolution
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The sum of the series `x+(2^(3))/(2!)x^(2)+(3^(3))/(3!)x^(3)+(4^(3))/(4!)x^(4)`+……..to `infty` isA. `(x+x^(2)+x^(3))e^(x)`B. `(x^(2)+x^(3))e^(x)`C. `(x+3x^(2)+x^(3))e^(x)`D. `x^(3)e^(x)` |
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Answer» Answer: We have `(x+2^(3))/(2!)x^(2)(3^(3))/(3!)x^(3)+(4^(3))/(4!)x^(4)+….` `=underset(n=1)overset(infty)Sigma (n^(3))/(n!)x^(n)` `=underset(n=1)overset(infty)Sigma(n+3n(n-1)+n(n-1)(n-2))/(n!)x^(n)` `underset(n=1)overset(infty)Sigma{(n)/(n!)+(3n(n-1))/(n!)+(n(n-1)(n-2))/(n!)}x^(n)` `=underset(n=1)overset(infty)Sigma(x^(n))/(n-1)!+3underset(n=1)overset(infty)Sigma(x^(n))/(n-2)! underset(n=1)overset(infty)Sigma(x^(n))/(n-3)!` `=xe^(x)+3x^(2)e^(x)+x^(3)e^(x)=(x+3x^(2)+x^(3))e^(x)` |
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