Given n = 3, evaluate \(\frac{1}{(n-1)!} - \frac{1}{(n+1)!}\)
Question 1 Report
Given n = 3, evaluate \(\frac{1}{(n-1)!} - \frac{1}{(n+1)!}\)
Answer Details
We are given that n=3. Substituting n=3, we have:
\(\frac{1}{(n-1)!} - \frac{1}{(n+1)!} = \frac{1}{(3-1)!} - \frac{1}{(3+1)!} = \frac{1}{2!} - \frac{1}{4!}\)
Evaluating the factorials, we get:
\(\frac{1}{2!} - \frac{1}{4!} = \frac{1}{2} - \frac{1}{24} = \frac{12}{24} - \frac{1}{24} = \frac{11}{24}\)
Therefore, the value of \(\frac{1}{(n-1)!} - \frac{1}{(n+1)!}\) when n=3 is \(\frac{11}{24}\).
Hence, the answer is, \(\frac{11}{24}\).
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