Need Help Finishing Inductive Proof Containing Subscripts

Question

The Proposition is as follows: Construct a sequence of numbers a₁ , a₂, a₃, a₄ as follows. Set a₁ = 5 then triple each term of the sequence and subtract four to obtain the next term. Prove the n^th term is given by the forumula a_n = 3ⁿ + 2.

MY WORK SO FAR: I wrote out some of the sequence: 5, 11, 29, 83 I then created the following statement: S_n = 5, 11, 29, 83... 3 (a_{n - 1}) - 4 = 3ⁿ + 2 ; as long as n >= 2.

Base Case: n = 2, in the left hand side we get 11 and in the right hand side we get 11.

Inductive Step: fix k: S_k = 5, 11, 29, 83... 3 (a_{k - 1}) - 4 = 3^k + 2

S_k+1 = 3(_{ak + 1 - 1}) - 4 = 3^{k + 1} + 2.

I then tried to put it into the following equation S_k + a_k = S_{k + 1} But I never can get it to properly equal one another.

All Help would be much appreciated as I'm not looking for a direct answer, rather a helpful explanation and hints towards a correct answer.

Again I am not looking for direct answers

Answer 1

The question is all about the sequence $a_1, a_2, a_3, \ldots .$ There is no mention of taking any sum of these terms; only a rule so that if you know one of the terms (such as $a_{15}$) you can compute the next term (so if you know $a_{15}$ you can compute $a_{16}$).

When they write, "Prove the $n$th term is given by ...," they mean "prove that $a_n$ is ... ."

For example, the first term is $a_1 = 5,$ and indeed when $n = 1$ we see that $$3^n + 2 = 3^1 + 2 = 3 + 2 = 5.$$

But if $n = 2$ then $$3^n + 2 = 3^2 + 2 = 9 + 2 = 11,$$ which we see is the second term of the sequence. And if $n = 3$ then $$3^n + 2 = 3^3 + 2 = 27 + 2 = 29,$$ which is the third term.

If you try to prove that the sum of the first $n$ terms is $3^n + 2,$ you will not succeed, because that theorem is false.

The statement we are actually supposed to prove is that $a_n = 3^n + 2.$ Notice we have already shown it to be true for $n = 1,$ so you can use $n = 1$ rather than $n = 2$ for your base case. For the inductive case you will only need to look at $a_k$ and $a_{k+1}.$

Answer 2

"Set a1 = 5 then triple each term of the sequence and subtract four to obtain the next term." So $a_1= 5$ and $a_{n+1}= 3a_n- 4$. You want to use induction to prove that $a_n= 3^n+ 2$. When n= 1, $a_1= 3^1+ 2= 3+ 2= 5$ is true.

Now the induction step: Suppose that, for some k, $a_k= 3^k+ 2$. Then $a_{k+1}= 3a_k- 4= 3(3^k+ 2)- 4= 3^{k+1}+ 6- 4= 3^{k+1}+ 2$. QED.