If $(n+1)!*f(n) = (n-1)!$ , what is the value of $f(1) + f(2) + … + f(100)$ ?

Correct answer marked below

\frac{1}{99}

\frac{1}{100}

\frac{1}{101}

\frac{99}{100}

\frac{100}{101}

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Explanation

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Official Explanation

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(n+1)!*f(n) = (n-1)!

tells us that

f(n) = \frac{(n-1)!}{(n+1)!}

f(1)=\frac{(n-1)!}{(n+1)!}=\frac{0!}{2!}=\frac{1}{2}

.

To that, we are going to add more positive numbers, so we can eliminate answer choices A, B, and C.

The terms will be:

\frac{(n-1)!}{(n+1)!}+\frac{(n)!}{(n+2)!}+\frac{(n+1)!}{(n+3)!}+\frac{(n+2)!}{(n+4)!}. . .

…

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What this question tests

Your ability to evaluate and compose functions, including unusual defined operations.

Applying function operations in the wrong order or ignoring domain restrictions.