InterviewSolution
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InterviewSolution
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The value of `9+(16)/(2!)+(27)/(3!)+(42)/(4!)+……infty` isA. `9e-6`B. `11e-6`C. `13e-6`D. `12e-6` |
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Answer» Answer: Consider the series 9+16+27+42+…. We observe that the successive differences of the terms of this series are 7,11,15… Clearly these are in A.P so let its `n^(th)` term be `t_(n)=an^(2)+bn+c` `rarr t_(1)=a+b+c rarr a+b+c=9` `t_(2)=4 a+2b+c rarr 4a+2b+c=16` `t_(3)=9a+3b+c rarr 9a+3b+c=27` Solving these equation we get a=2,b=1 and c =6 `therefore t_(n)=2n^(2)+n+6` Thus we have `9+(16)/(2!)+(27)/(3!)+(42)/(4!)+...infty` `=underset(n=1)overset(infty)Sigma(2n^(2)+n+6)/(n!)` `=2underset(n=1)overset(infty)Sigma(n^(2))/(n!)+underset(n=1)overset(infty)Sigma(n)/(n!)+6 underset(n=1)overset(infty)Sigma(1)/(n!)` =11e-6 |
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