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.
