Use Mathematical Induction to show that 512/8*5^(n)+6*3^(2n)-80n-14 for all n in N.
Answers
Step-by-step explanation:
We need to prove that 8(5)n+6(3)2n−80n−14 is divisible by 512 for all n∈N.
I started by taking n=1, then for n=k and then for n=k+1.
So in this question we have to prove that difference of nth and (n+1)th term is divisible by 512.
8(5n+1−5n)+6(32n+2−32n)−80(n+1−n)
when divided by 512 gives remainder 0. Can you continue it now OP?
Also the attempt which you showed in the question in that you cannot write n=n+1. You can write it like n=x+1 instead.
For simplicity, let f(n)=8(5)n+6(3)2n−80n−40. We can simplify this as
f(n)=8(5)n+2(3)2n+1−40(2n+1).
We are trying to prove that 512 divides f(n)−f(n−1). We see that it is
[8(5)n+2(3)2n+1−40(2n+1)]−[8(5)n−1+2(3)2n−1−40(2n−1)]8(5)n−8(5)n−1+2(3)2n+1−2(3)2n−1−40(2n+1)+40(2n−1)8(5)n−1(5−1)+2(3)2n−1(32−1)−8032(5)n−1+16(3)2n−1−8016(2(5)n−1+(3)2n−1−5)
Because we factored 16, we see that 512/16=32. Hence, we need to prove that 32 divides 2(5)n−1+(3)2n−1−5. Let this expression be g(n).
The condition n∈N tells us that the base case should be n=1. We see that g(1)=0 which is trivial.