For each \(n \in N, 3^{2 n}-1\) is divisible by (a) 8 (b) 16 (c) 32 (d) None of these

The given expression is \(3^{2n} - 1\). It is stated that this expression is divisible by either 8, 16, 32, or none of these.

First, let's examine the expression for different values of n. For n=1, \(3^{2(1)} - 1 = 3^2 - 1 = 9 - 1 = 8\) For n=2, \(3^{2(2)} - 1 = 3^4 - 1 = 81 - 1 = 80\) For n=3, \(3^{2(3)} - 1 = 3^6 - 1 = 729 - 1 = 728\) Now let's check for the factors of 8, 16, and 32 in the results. 8 | 8 8 | 80 8 | 728 From the above, we can see that all results are divisible by 8. Now, let's try to generalize it by using algebra.

Consider the expression \(3^{2n} - 1\). We can rewrite it as follows, using the formula \(a^{2n} = (a^n)^2\): \(3^{2n} - 1 = (3^n)^2 - 1\) Now, notice the resemblance between this expression and the formula for the difference of squares: \(a^2 - b^2 = (a + b)(a - b)\). Let's use this formula to factor the given expression: \((3^n)^2 - 1^2 = (3^n + 1)(3^n - 1)\) As we can see, the given expression is factored into two factors: \((3^n + 1)\) and \((3^n - 1)\). Now let's analyze these two factors: \(3^n + 1\) has a pattern of odd numbers for all positive integer values of n. Additionally, notice that \(3^n - 1\) has a pattern of numbers that are divisible by 8. Now since one factor is odd and the other factor has a pattern of numbers divisible by 8, the product \((3^n + 1)(3^n - 1)\) should be divisible by 8. Therefore, we can conclude that for all \(n \in N\), the expression \(3^{2n} - 1\) is divisible by 8. #Answer#: (a) 8