InterviewSolution
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InterviewSolution
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The sum of the series `1+(2^(3))/(2!)+(3^(3))/(3!)+(4^(3))/(4!)`+… to `infty` isA. 2eB. 3eC. 4eD. 5e |
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Answer» Answer: We have `1+(2^(3))/(2!)+(3^(3))/(3!)+(4^(3))/(4!)+…underset(n=1)overset(infty)Sigma(n^(3))/(n!)` Let `n^(3)=a_(0)+a_(1)n+a_(2)n(n-1)+a+_(3)n(n-1)(n-2)` By comparing the coefficients of like powers of n on both sides we get Substituting the value in (i) we get `n^(3)=n+3n(n-1)+n((n-1)(n-2)` `therefore 1+(2^(3))/(2!)+(3^(3))/(3!)+(4^(3))/(4!)+...` `-underset(n=1)overset(infty)Sigma(n^(3))/(n!)` `=underset(n=1)overset(infty)Sigma(n+3n(n-1)+n(n-1)(n-2))/(n!)` =e+3e+e=5e |
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