91Ó°ÊÓ

Let's first verify the base case for \(n = 1\). The given sequence, in this case, is just \((2)\). According to the formula, it should equal \(\frac{(2(1))!}{1!}\). So, let's check it: Given sequence: \((2)\) Formula: \(\frac{(2(1))!}{1!} = \frac{2!}{1!} = \frac{2}{1} = 2\) As we can see, both the given sequence and the formula give the same result for the base case of \(n = 1\).

We will assume that the given formula is true for an arbitrary natural number \(n = k\): \((2)(6)(10)(14) \cdots(4k-2) = \frac{(2k)!}{k!}\)

Now we need to prove that the formula is true for \(n=k+1\): \((2)(6)(10)(14) \cdots(4k-2)(4(k+1)-2)\) Using our assumption \((2)(6)(10)(14) \cdots(4k-2) = \frac{(2k)!}{k!}\), we can rewrite the expression as: \(\frac{(2k)!}{k!}(4(k+1)-2) = \frac{(2k)!}{k!}(4k+2)\) Our goal is to show that this expression is equal to \(\frac{(2(k+1))!}{(k+1)!}\): \(\frac{(2(k+1))!}{(k+1)!} = \frac{(2k+2)!}{(k+1)!}\) To prove this, let's manipulate the given expression \(\frac{(2k)!}{k!}(4k+2)\): \(\frac{(2k)!}{k!}(4k+2) = \frac{(2k)!(4k+2)}{k!}\) Now, we can rewrite \((2k+2)!\) as \((2k+2)(2k+1)(2k)!\): \(\frac{(2k)!(4k+2)}{k!} = \frac{(2k+2)(2k+1)(2k)!}{k!}\) To transform the denominator into \((k+1)!\), we multiply and divide by \((k+1)\): \(\frac{(2k+2)(2k+1)(2k)!}{k!} = \frac{(2k+2)(2k+1)(2k)!}{(k+1)k!} \cdot \frac{(k+1)}{(k+1)}\) By doing this, we get: \(\frac{(2k+2)(2k+1)(2k)!}{(k+1)k!} \cdot \frac{(k+1)}{(k+1)} = \frac{(2k+2)(2k+1)(2k)!}{(k+1)!}\) As we can see, both expressions are equal: \(\frac{(2k+2)(2k+1)(2k)!}{(k+1)!} = \frac{(2(k+1))!}{(k+1)!}\)

Since we successfully proved the formula for the base case when \(n=1\) and verified the formula for \(n=k+1\) using mathematical induction, we can now conclude that the given formula \((2)(6)(10)(14) \cdots(4 n-2)=\frac{(2 n) !}{n !}\) holds true for all natural numbers \(n\).