Prove 2^(n+2) + 3^(2n+1) is a multiple of 7 for all positive integers of n by mathematical induction.

Let P(n) be the proposition that 2ⁿ⁺² + 3²ⁿ⁺¹ is a multiple of 7 for all positive integers of n.
Let n=12³ + 3³ = 8 + 27 = 35 = 7(5)This is divisible by 7.
Assume n=k2^k+2 + 3^2k+1 = 7m
The above equation can be rearranged to 2^k+2 = 7m - 3^2k+1, which will become useful later.
Test n=k+12^(k+1)+2 + 3^2(k+1)+12^k+3 + 3^2k+32(2^k+2)+ 3^2k+32(7m - 3^2k+1)+ 3^2k+3 The above step is done using the rearrangement of the equation from the 'assume n=k' section. 14m - 2(3^2k+1) + 9(3^2k+1)14m + 7(3^2k+1)7(2m + 3^2k+1)The above is divisible by 7.
As P(1) was shown to be true, and when n=k was assumed true, P(k+1) was proven true, P(n) has been proven true for all positive integers of n by the principle of mathematical induction.