I’m assuming the task is to prove that if n>=0 (where n is a whole number) then 4^(2n+1)+3^(n+2) is a multiple of 13.
Let’s prove this using induction:
Base case: Let n=0, we want to show that LHS=4^(2(0)+1)+3^(0+2)=13k=RHS where kinNN
LHS:
4^(2(0)+1)+3^(0+2)=4^1+3^2=4+9=13
RHS:
Let k=1, so 13k=13(1)=13 Thus the RHS = LHS.
Inductive step:
Assume that 4^(2n+1)+3^(n+2)=13j where jinNN
We want to show that LHS=4^(2(n+1)+1)+3^((n+1)+2)=13h=RHS where hinNN
LHS:
4^(2(n+1)+1)+3^((n+1)+2)=4^(2n+1+2)+3^(n+2+1)
Use the product rule:
4^(2n+1)4^2+3^(n+2)3^1=16*4^(2n+1)+3*3^(n+2)
Use some clever rewriting:
(13+3)*4^(2n+1)+3*3^(n+2)=13*4^(2n+1)+3*4^(2n+1)+3*3^(n+2)
After factoring out the 3, we can use the inductive hypothesis to rewrite the LHS so that:
13*4^(2n+1)+3(4^(2n+1)+3^(n+2))=13*4^(2n+1)+3(13j)=13(4^(2n+1)+3j)
Let m=4^(2n+1) where m inNN
Then 13m+13(3j)=13(m+3j)
Let m+3j=h (which is okay because a natural number plus a natural number is another natural number)
Then LHS = 13h=RHS.
This proves the inductive step which completes the proof.
