MHB Arithmetic Concepts: Get Help Understanding Problems

Click For Summary
The discussion focuses on solving a recursive arithmetic problem using the formula $a_n = 3a_{n-1} + 2$. The user seeks clarification on calculating terms in the sequence, starting with $a_2$ and $a_3$, based on the initial value $a_1 = 1$. Another participant highlights the difference between recursive and closed-form problems, providing an example for clarity. The conversation emphasizes the importance of understanding the structure of the problem to find solutions effectively. Overall, the thread serves as a collaborative effort to assist with arithmetic concepts and problem-solving strategies.
Coder74
Messages
20
Reaction score
0
Hi everyone!
I'm having trouble solving this problem, its set up in a way that I don't understand and I was hoping someone could help clarify it with me..Thanks a bunch!

View attachment 6475
 

Attachments

  • Capture.PNG
    Capture.PNG
    13.2 KB · Views: 101
Mathematics news on Phys.org
Hi Coder74,

Let's first find $a_2$. To do so, set $n = 2$ in the recursion formula $a_n = 3a_{n-1} + 2$ to get $a_2 = 3a_1 + 2$. It's given that $a_1 = 1$, so $a_2 = 3(1) + 2 = 5$. Now consider $a_3$. Set $n = 3$ in the recursion formula to get $a_3 = 3a_2 + 2 = 3(5) + 2 = 17$. Continue to $a_5$.
 
Thanks so much!
I have a similar question but this doesn't have any terms to be substituted like the above equation.. How do I go about starting this? I've been sitting here like a bump on a log for a while and my teacher has gone offline..(I'm homeschooled)
I really appreciate everyone and their hard work on this website!

Example: View attachment 6481

This is a practice question I sampled from the extra help section..it looks exactly like the one I'm having trouble with but it looked a little easier.
 

Attachments

  • sHelp.PNG
    sHelp.PNG
    13 KB · Views: 94
In this example, you are given the closed form, rather than a recursion, so you simply need to use $n=10$, that is:

$$a_{10}=2(-1)^{10}=?$$

In the future, please begin a new thread for a new problem. :D
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

Similar threads

Replies
13
Views
2K
Replies
28
Views
646
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
4K
  • · Replies 33 ·
2
Replies
33
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K