Prove by mathematical induction that 2^(2n-1) + 3^(2n-1) is divisible by 5 for all natural numbers n.

First check that this works for n=1:2^(2x1 - 1) + 3^(2x1 - 1) = 2^1 +3^1 = 5 (so true for n=1)Now we assume this to work for any n = k.Assumption: 2^(2k-1) + 3^(2k-1) = 5a, where a is some integer constant.Now we check that this works for n = k + 1:2^(2(k+1)-1) + 3^(2(k+1)-1) (we try to manipulate this algebra so that we can get it in the form 5a)=2^(2k+2-1) + 3^(2k+2-1) = 2^2 x 2^(2k-1) + 3^2 x 3^(2k-1) = 4 x 2^(2k-1) + 9 x 3^(2k-1) = 9(2^(2k-1) + 3^(2k-1)) - 5(2^(2k-1)) (notice that we have our assumption, which we can write as 5a)= 9(5a) - 5(2^(2k-1)) = 5(9a - 2^(2k-1)) (9a - 2^(2k-1) is some integer constant, we can write this as b)=5bThis is true for n=1. If it is true for n=k, then we have shown it to be true for n=k+1 also. Therefore by mathematical induction it is true for all positive integers n.