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WAEC Further Mathematics 2016 Paper

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Question : 14

Total: 40

Given n = 3, evaluate

\;\frac{1}{(n-1)!} - \frac{1}{(n+1)!}

$12$
$2\frac{1}{2}$
$2$
$\;\frac{11}{24}$

Solution:

Concept:

Substitute the given value of

n

into the factorial expression.

Explanation:

Given

n = 3

, substitute into

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

.

This gives

\frac{1}{(3-1)!} - \frac{1}{(3+1)!}

.

So we have

\frac{1}{2!} - \frac{1}{4!}

.

Now

2! = 2

and

4! = 24

.

Therefore,

\frac{1}{2} - \frac{1}{24}

.

Convert to a common denominator:

\frac{12}{24} - \frac{1}{24} = \frac{11}{24}

.

Answer:

\frac{11}{24}

Correct option: D.

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