If ( \frac{(2n+2)!}{(2n)!}=132 ), what is the value of (n)?

Correct answer: (5)

The direct answer is D, \(n=5\). Expand the numerator until the denominator appears: \((2n+2)!=(2n+2)(2n+1)(2n)!\). Thus \(\frac{(2n+2)!}{(2n)!}=(2n+2)(2n+1)=132\). Among the options, if \(n=5\), the factors are \(2(5)+2=12\) and \(2(5)+1=11\), and \(12\times11=132\). Therefore D is correct. Option A, 3, gives \(8\times7=56\), not 132. Option B, 4, gives \(10\times9=90\), not 132. Option C, 6, gives \(14\times13=182\), which is too large. Option D, 5, matches exactly. The key is to cancel the common \((2n)!\) first and then solve the product of the two consecutive factors.