Olympiad Maths Prep

Track / Stage 4 / 142 of 340 #402 of 2000

Problem 402

AMC 12 late, AIME early
Algebra Difficulty 4.8 Find the answer

3. If Sn=n![12!+23!+⋯+n(n+1)!−1]S_{n}=n!\left[\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{n}{(n+1)!}-1\right], then
S2013= S_{2013}=

Official solution

3. −12014-\frac{1}{2014}.

From k(k+1)!=(k+1)−1(k+1)!=1k!−1(k+1)!\frac{k}{(k+1)!}=\frac{(k+1)-1}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}, we know
12!+23!+⋯+n(n+1)!=1−11!+2−12!+3−13!+⋯+(n+1)−1(n+1)!=1−1(n+1)!. \begin{array}{l} \frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{n}{(n+1)!} \\ =\frac{1-1}{1!}+\frac{2-1}{2!}+\frac{3-1}{3!}+\cdots+\frac{(n+1)-1}{(n+1)!} \\ =1-\frac{1}{(n+1)!} . \end{array}

Therefore, Sn=n!{[1−1(n+1)!]−1}=−1n+1S_{n}=n!\left\{\left[1-\frac{1}{(n+1)!}\right]-1\right\}=-\frac{1}{n+1}.
Thus, S2013=−12014S_{2013}=-\frac{1}{2014}.

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