Please use mathematical induction to solve: prove: 3 ^ 2n-2 ^ 2n can be divided by 5 (n belongs to positive natural number)

1. N = 13 & # 178; - 2 & # 178; = 9-4 = 5 obviously holds. 2. N = k holds when 3 ^ 2k-2 ^ 2K is a multiple of 5. When n = K + 1, 3 ^ 2 (K + 1) - 2 ^ 2 (K + 1) = 9 × 3 ^ 2k-4 × 2 ^ 2K = 9 × 3 ^ 2k-9 × 2 ^ 2K + 5 × 2 ^ 2K = 9 × (3 ^ 2k-2 ^ 2K) + 5 × 2 ^ 2K, because 3 ^ 2k-2 ^ 2K is a multiple of 5, and 5 × 2 ^ 2K is also

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