Maths Olympiad Prep

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Algebra Difficulty 5.0 AIME, harder Find the answer

24 Calculate 31!+2!+3!+42!+3!+4!+⋯+n+2n!+(n+1)!+(n+2)!\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+\cdots+\frac{n+2}{n!+(n+1)!+(n+2)!}

\qquad

A number or a short expression. Spacing and $ signs are ignored.

Solution

2412−1(n+2)!24 \frac{1}{2}-\frac{1}{(n+2)!}. From k!+(k+1)!+(k+2)!=k!(k+2)2k!+(k+1)!+(k+2)!=k!(k+2)^{2}, we have k+2k!+(k+1)!+(k+2)!=1k!(k+2)=1(k+1)!−1(k+2)!\frac{k+2}{k!+(k+1)!+(k+2)!}=\frac{1}{k!(k+2)}=\frac{1}{(k+1)!}-\frac{1}{(k+2)!}. Therefore, the original expression == (12!−13!)+(13!−14!)+⋯+(1(n+1)!−1(n+2)!)=12−1(n+2)!\left(\frac{1}{2!}-\frac{1}{3!}\right)+\left(\frac{1}{3!}-\frac{1}{4!}\right)+\cdots+\left(\frac{1}{(n+1)!}-\frac{1}{(n+2)!}\right)=\frac{1}{2}-\frac{1}{(n+2)!}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.