InterviewSolution
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InterviewSolution
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If `a_n = n (n!)`, then `sum_(r=1)^100 a_r` is equal toA. 101!B. 100!-1C. 101!-1D. 101!+1 |
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Answer» We have, `sum_(r=1)^(100)a_(r)=sum_(r=1)^(100)r(r!)=sum_(r=1)^(100){(r+1)!-1}r!` `impliessum_(r=1)^(100)a_(r)=sum_(r=1)^(100){(r+1)!-r!}` `impliessum_(r=1)^(100)a_(r)=(2!-1!)+(3!-2!)+(4!-3!)+......+(101!-100!)` `impliessum_(r=1)^(100)a_(r)=101!-1` |
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