Use iteration to make an educated guess at an explicit formula

[Topgun_68]

Homework Statement

The sequence {e_n} is defined recursively as e₀ = 3; e_k = 4e_k-1 + 7, for all k ≥ 1. Use iteration to make an educated guess at an explicit formula for the sequence.

The attempt at a solution

I spent all day on this one and I'm still lost.

e₁ = 4 x e₀ + 7
= 4 x 3 + 7 {since e₀ = 3}
= 19

e₂ = 4 x e₁ + 7
= 4 x 19 + 7 {since e₁ = 19}
= 83

e₃ = 4 x e₂ + 7
= 4 x 83 + 7 {since e₂ = 83
= 339

I came up with this, and it looks like a geometric sequence, but I can't figure out how to create the formula using the using Theorem 5.2.3 {Sum of a Geometric Sequence}

3 + 4² + 4³ + 4⁴ ... 4ⁿ⁺¹

Thanks for any guidance!]

 

[Curious3141]

Homework Statement

The sequence {e_n} is defined recursively as e₀ = 3; e_k = 4e_k-1 + 7, for all k ≥ 1. Use iteration to make an educated guess at an explicit formula for the sequence.

The attempt at a solution

I spent all day on this one and I'm still lost.

e₁ = 4 x e₀ + 7
= 4 x 3 + 7 {since e₀ = 3}
= 19

e₂ = 4 x e₁ + 7
= 4 x 19 + 7 {since e₁ = 19}
= 83

e₃ = 4 x e₂ + 7
= 4 x 83 + 7 {since e₂ = 83
= 339

I came up with this, and it looks like a geometric sequence, but I can't figure out how to create the formula using the using Theorem 5.2.3 {Sum of a Geometric Sequence}

3 + 4² + 4³ + 4⁴ ... 4ⁿ⁺¹

Thanks for any guidance!]

Yup, you're on the right track. Remember that the geometric series only begins from $4^2$, and you have to add that 3 on to any sum you get (the "3" is not part of the geometric series).

What's the sum of n terms of a geometric series that begins like this: $4^2, 4^3, 4^4,...4^{n+1}$? (that's actually n terms)

 

[Topgun_68]

Would it be (4^(n-1)-1) / 3

Can't get superscript & frac to work at the same time

Yup, you're on the right track. Remember that the geometric series only begins from $4^2$, and you have to add that 3 on to any sum you get (the "3" is not part of the geometric series).

What's the sum of n terms of a geometric series that begins like this: $4^2, 4^3, 4^4,...4^{n+1}$? (that's actually n terms)

 

[Curious3141]

Would it be (4^(n-1)-1) / 3

Can't get superscript & frac to work at the same time

You're missing a factor of 16 there, and your exponent is off by one). The geometric series starts with $4^2 = 16$, so its sum to n terms is $\displaystyle \frac{4^2(4^n-1)}{3}$. All you have to do now is to add 3 to that expression and simplify it to get a neater expression and just verify that the expression works for a few terms. I think you can stop there for the purposes of this question, but to be rigorous, you should prove that expression by induction.

 

[Topgun_68]

Thanks for the help. I simplified it to this:

$\displaystyle \frac{(4^{n+2}-7)}{3}$

I never realized their was so many ways that I could represent the same equation. The next part of my question is to prove it using mathematical induction so I want to make sure I have the simplest formula to do so. Let me know if their is a simpliar formula to prove it but I am working on this now. Thanks again for everyones help!

You're missing a factor of 16 there, and your exponent is off by one). The geometric series starts with $4^2 = 16$, so its sum to n terms is $\displaystyle \frac{4^2(4^n-1)}{3}$. All you have to do now is to add 3 to that expression and simplify it to get a neater expression and just verify that the expression works for a few terms. I think you can stop there for the purposes of this question, but to be rigorous, you should prove that expression by induction.

 

[Curious3141]

Thanks for the help. I simplified it to this:

$\displaystyle \frac{(4^{n+2}-7)}{3}$

I never realized their was so many ways that I could represent the same equation. The next part of my question is to prove it using mathematical induction so I want to make sure I have the simplest formula to do so. Let me know if their is a simpliar formula to prove it but I am working on this now. Thanks again for everyones help!

That's pretty much the simplest formula, and it's very easy to do the inductive step using that formula.

 

[Topgun_68]

Sorry to bring this post back up, but I'm not finding it so easy to do the inductive step. Can you or someone else let me know what I'm doing wrong.

A: (Basic Step)

Prove for n = 0:

e₀= $\displaystyle \frac{(4^{0+2}-7)}{3}$ = 3 (Initital condition)

B: Assume n = k

e_k= $\displaystyle \frac{(4^{k+2}-7)}{3}$ for some integer k

C: Prove for n = k+1

e_(k+1) = 4e_(k+1)-1 + 7

e_(k+1) = 4e_k + 7

e_(k+1) = $\displaystyle 4 \frac{(4^{k+2}-7)}{(3)}$ +7 (replaced k by the Inductive Hypothesis)

$\displaystyle \frac{(4^{k+3}-28)}{(3)}$

I can't seem to get it back to the explicit equation to complete the proof.

That's pretty much the simplest formula, and it's very easy to do the inductive step using that formula.

 

[Curious3141]

Sorry to bring this post back up, but I'm not finding it so easy to do the inductive step. Can you or someone else let me know what I'm doing wrong.

A: (Basic Step)

Prove for n = 0:

e₀= $\displaystyle \frac{(4^{0+2}-7)}{3}$ = 3 (Initital condition)

B: Assume n = k

e_k= $\displaystyle \frac{(4^{k+2}-7)}{3}$ for some integer k

C: Prove for n = k+1

e_(k+1) = 4e_(k+1)-1 + 7

e_(k+1) = 4e_k + 7

e_(k+1) = $\displaystyle 4 \frac{(4^{k+2}-7)}{(3)}$ +7 (replaced k by the Inductive Hypothesis)

$\displaystyle \frac{(4^{k+3}-28)}{(3)}$

I can't seem to get it back to the explicit equation to complete the proof.

e_(k+1) = $\displaystyle 4 \frac{(4^{k+2}-7)}{3}$ +7 (replaced k by the Inductive Hypothesis)

Right up to here. (I removed the brackets around the denominator 3, because you don't have to keep putting them there).

The next line should be $\displaystyle \frac{4^{k+3}-28}{3} + 7$ (you're missing the "plus 7" in yours).

Now express $7 = \frac{21}{3}$, and sum it into that expression with the same denominator (3). Simplify.

 

[Topgun_68]

Dam, I forgot to drop the +7 down on my notes. Either way I didn't even think to express +7 as

\frac{21}{3} . I keep learning something new every time I post to these forums.

Thanks Everyone!

e_(k+1) = $\displaystyle 4 \frac{(4^{k+2}-7)}{3}$ +7 (replaced k by the Inductive Hypothesis)

Right up to here. (I removed the brackets around the denominator 3, because you don't have to keep putting them there).

The next line should be $\displaystyle \frac{4^{k+3}-28}{3} + 7$ (you're missing the "plus 7" in yours).

Now express $7 = \frac{21}{3}$, and sum it into that expression with the same denominator (3). Simplify.

 

[Curious3141]

Thanks Everyone!

"Everyone" says "You're welcome".

 

[Topgun_68]

Ha, I thought someone else posted in here too, but that was a different thread. Thanks to just you! Everyone else who will learn from this thread thanks you too!

"Everyone" says "You're welcome".