Prove, by induction, that 4^(n+1) + 5^(2n-1) is always divisible by 21

Firstly when proving something by induction, we always show that the base case works, i.e. we plug in n=1. In this case we get 4² + 5¹ =21, which is divisible by 21. Next we state what is called the inductive hypothesis, this just means we assume it is true for n=k, i.e. we assume 4^k+1 + 5^2k-1 is divisible by 21. Now we try to use this to show that the case n=k+1 is also true, in this case we consider 4^k+2 + 5^2(k+1)-1 which can be manipulated as shown below4^k+2 +5^2k+1 = 4* 4^k+1 + 55 5^2k-1 = 4* 4^k+! + 4* 5^2k-1 + 21* 5^2k-1 = 4*( 4^k+1 + 5^2k-1) + 21* 5^2k-1Now, as we know that 4^k+1 + 5^2k-1 is divisible by 21, we can see that the expression above is also divisible by 21.Lastly we conclude that because the case n=1 is true and if n=k is true then n=k+1 is true, the statement is true for all natural numbers n.