InterviewSolution
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InterviewSolution
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The sum of the series `(1^(2))/(1!)+(2^(2))/(2!)+(3^(2))/(3!)`+……to `infty` isA. eB. 2eC. 3eD. none of these |
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Answer» Answer: We have `(1)^(2)/(1!)+(2)^(2)/(2!)+(3)^(2)/(3!)+…….underset(n=1)overset(infty)Sigma(n)^(2)/(n!)` Let `n^(2)=a_(0)+a_(1)=a_(2)=1` By comparing the coefficeints of like poweres of n both sides we get `a_(0)=0,a_(1)-a_(2)=0 and a_(2)=1` `rarr a_(0)=0, a_(1)=a_(2)=1` Putting the value of `a_(0,a_(1),a_(2)` (i) we get `n^(2)=n+n(n-1)` `therefore underset(n=1)overset(infty)Sigma(n^(2))/(n=1)(n+n(n-1))/(n!)` `rarr underset(n=1)overset(infty)Sigma(n^(2))/(n=1){(n)/(ni+n(n-1))/(ni)}` |
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