InterviewSolution
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InterviewSolution
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The sum of the series `1+(2^(4))/(2!)+(3^(4))/(3!)+(4^(4))/(4!)+(5^(4))/(5!)`+…..isA. 12eB. 5eC. 14eD. 15e |
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Answer» Answer: We have `1+(2^(4))/(2!)+(3^(4))/(3!)+(4^(4))/(4!)+(5^(4))/(5!)+…underset(n=1)overset(infty)Sigma (n^(4))/(n!)` Let `n^(4)=a_(0)+a_(1)n+a_(2)n(n-1)` `+a_(3)n(n-1)(n-2)+a_(4)n(n-1)(n-32)(n-3)` On equating the coefficients of like poweres of n on both sides we get `a_(0)=0,a_(1)-a_(2)+2a_(3)-6a_(4)=0,a_(2)-3a_(3)+11a_(4)=0a_(3)-6a_(4) and a_(4)=1` `rarr a_(0=0a_(1)=1 , a_(2)=7 , a_(3)=6,a_(4)=1` substituting the value of `a_(0),a_(1),a_(2),a_(3),a_(4)` (i) we get `n^(4)=n+7n(n-1)+6n(n-1)(n-2)+n(n-21)+n(n-1)(n-2)(n-3)` `therefore 1+(2^(4))/(2!)+(3^(4))/(3!)+(4^(4))/(4!)`+... `=underset(n=1)overset(infty)Sigma(1)/(n-1)!+7 underset(n=2)overset(infty)Sigma(1)/(n-2)!+6 underset(n=3)overset(infty)Sigma(1)/(n-3)!+underset(n=4)overset(infty)Sigma(1)/(n-3)!+underset(n=4)overset(infty)Sigma(1)/(n-4)!` `=e+7e+6e+e=15e` |
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