Interesting Sequence (Not Arithmetic or Geometric)

[Scrutinize]

Gyazo

gyazo.com

I know how to do the first part, I substituted an+1 and got 2 for the answer (Which is correct)

This question is a geometric and arithmetic sequence, and I'm not sure how to do part b and c. I tried using both arithmetic and geometric equations to see if i can see a similarity and work of that but I can't find anything. On the test just to note, I'll only have arithmetic and geometric rules, so for whoever helps me please keep that in mind so that you know that I can only work of arithmetic and geometric rules.

Thank you so much! Any help would be very helpful.

 

[Steven G]

Your 1st step, in my opinion, is to write out the 1st few terms using the formula from part b and see why it is the same as the given sequence. Let us know how you make out.

 

[MarkFL]

a) Find the numeric value of the quotient:

[MATH]\frac{a_{n+1}+3}{a_n+3}[/MATH]
Okay, as you did, I would use the given rule to state:

[MATH]\frac{a_{n+1}+3}{a_n+3}=\frac{(2a_n+3)+3}{a_n+3}=\frac{2(a_n+3)}{a_n+3}=2\quad\checkmark[/MATH]
b) Show that [MATH]a_n=2^{n-1}(c+3)-3[/MATH]
I would write the given rule in the form:

[MATH]a_{n+1}-2a_n=3[/MATH]
The root of the homogeneous equation is \(r=2\), and so the homogeneous solution is:

[MATH]h_n=c_12^n[/MATH]
The particular solution must take the form:

[MATH]p_n=A[/MATH]
Substituting this into the difference equation, we obtain:

[MATH]A-2A=3\implies A=-3[/MATH]
And so the general solution is:

[MATH]a_n=c_12^n-3[/MATH]
Using the given initial value, we can determine the parameter:

[MATH]a_1=c_12^1-3=c\implies c_1=\frac{c+3}{2}[/MATH]
Hence:

[MATH]a_n=\frac{c+3}{2}2^n-3=2^{n-1}(c+3)-3[/MATH]
c) Given that \(c=7\), find a formula for [MATH]S_n=\sum_{i=1}^na_i[/MATH] in terms of \(n\) only.

Okay, I would write:

[MATH]S_n=\sum_{i=1}^n(2^{i-1}(c+3)-3)=(c+3)\sum_{i=1}^n(2^{i-1})-3\sum_{i=1}^n(1)=?[/MATH]
Can you complete the calculation and then substitute \(c=7\)?

 

[Dr.Peterson]

What I would do is to observe that [MATH]\frac{a_{n+1}+3}{a_n+3}= 2[/MATH] tells us that the sequence [MATH]b_n = a_n+3[/MATH] is a geometric sequence, with common ratio 2.

So, using only what you know about geometric sequences, you can write a formula for [MATH]b_n[/MATH], and then [MATH]a_n = b_n - 3[/MATH].

I strongly suspect, given what you have said about your context, that something like this is what you are expected to do.

 

[Steven G]

Yes, while driving today I realized that you result from part a shows that you do have a geometric sequence. Why did you think otherwise?

 

[Scrutinize]

What I would do is to observe that [MATH]\frac{a_{n+1}+3}{a_n+3}= 2[/MATH] tells us that the sequence [MATH]b_n = a_n+3[/MATH] is a geometric sequence, with common ratio 2.

So, using only what you know about geometric sequences, you can write a formula for [MATH]b_n[/MATH], and then [MATH]a_n = b_n - 3[/MATH].

I strongly suspect, given what you have said about your context, that something like this is what you are expected to do.

What makes it a geometric sequence with common ratio 2? And how did you get an = bn-3?

Confused about that.

 

[Dr.Peterson]

What makes it a geometric sequence with common ratio 2? And how did you get an = bn-3?

Confused about that.

Look at the ratio. It says that [MATH]a_n+3[/MATH] doubles each time; so this new sequence (not the original) is a geometric sequence. So I gave it a name, [MATH]b_n[/MATH].

Then, if [MATH]b_n = a_n + 3[/MATH], we can solve for [MATH]a_n[/MATH] and get [MATH]a_n = b_n - 3[/MATH].

Did you do what Jomo suggested in post #2? That's always the best way to approach a problem that is completely unfamiliar to you: play with it to get a feel for what is going on. If we take [MATH]a_1 = 7[/MATH], then [MATH]a_2 = 2(7)+3 = 17[/MATH], [MATH]a_3 = 2(17)+3 = 37[/MATH], ... . Now if you add 3 to each of these numbers, we get [MATH]b_1 = 10[/MATH], [MATH]b_2 = 20[/MATH], [MATH]b_3 = 40[/MATH], ..., and you can see that it doubles each time: a geometric sequence with common ratio 2.

 

[Scrutinize]

I got it thanks very much!