InterviewSolution
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InterviewSolution
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If `a_n=n/((n+1)!)` then find `sum_(n=1)^50 a_n`A. `(50!-1)/(50!)`B. `(51!-1)/(51!)`C. `(1)/(2(n-1)!)`D. none of these |
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Answer» We have, `a_(n)+(n)/((n+1)!)` `impliesa_(n)=((n+1)-1)/((n+1)!)` `impliesa_(n)=(1)/(n!)-(1)/((n+1)!)` `:.sum_(n=1)^(50)a_(n)=sum_(n=1)^(50){(1)/(n!)-(1)/(((n+1)!))}` `impliessum_(n=1)^(50)a_(n)=((1)/(1!)-(1)/(2!))+((1)/(2!)-(1)/(3!))+((1)/(3!)-(1)/(4!))+.....+((1)/(50!)-(1)/(51!))` `impliessum_(n=1)^(50)a_(n)=1-(1)/(51!)=(51!-1)/(51!)` |
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