InterviewSolution
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InterviewSolution
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Let `S_(k),k=1,2, . . . ,100`, denote the sum of the infinite geometric series whose first term is `(k-1)/(k!)` and the common ratio `(1)/(k)`. Then the value of `(100^(2))/(100!)+sum_(k=1)^(100) |(k^(2)-3k+1)S_(k)|`, isA. 3B. 6C. 8D. 9 |
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Answer» Correct Answer - A We have, `S_(k)=((k-1)/(k!))/(1-(1)/(k))=(1)/((k-1)!)` `:." "underset(k=2)overset(100)sum|(k^(2)-3k+1)S_(k)|` `=underset(k=2)overset(100)sum|{:({(k-1)^(2)-k})/((k-1)!):}|` `=underset(k=2)overset(100)sum|{:(k-1)/((k-2)!)-(k)/((k-1)!):}|` `=underset(k=2)overset(100)sum|{:((k-2)+1)/((k-2)!)-((k-1)+1)/((k-1)!):}|` `=underset(k=2)overset(100)sum|{:(1)/((k-3)!)-(1)/((k-2)!)+(1)/((k-2)!)-(1)/((k-2)!):}|` `=underset(k=2)overset(100)sum{(1)/((k-3)!)-(1)/((k-2)!)}underset(k=2)overset(100)sum{(1)/((k-2)!)-(1)/((k-1)!)}` `={(1-(1)/(1!))+((1)/(1!)-(1)/(2!))+((1)/(2!)-(1)/(3!))+ . . . +((1)/(97!)-(1)/(98!))}` `+{(1-(1)/(1!))+((1)/(1!)-(1)/(2!))+((1)/(2!)-(1)/(3!))+ . . . +((1)/(98!)-(1)/(99!))}` `=(1-(1)/(98!))+(1-(1)/(99!))=2(1)/(98!)-(1)/(99!)=2-(100)/(99!)` Hence, `((100)^(2))/(100!)+underset(k=1)overset(100)sum|{:(k^(2)-3k+1)S_(k):}|` `(100)/(99!)+S_(1)+underset(k=2)overset(100)sum|{:(k^(2)-3k+1)S_(k):}|` `=(100)/(99!)+1+2-(100)/(99!)=3` |
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