Recursion Formula: Solve Series with Ease

  • Thread starter Thread starter Poweranimals
  • Start date Start date
  • Tags Tags
    Formula Recursion
AI Thread Summary
The recursion formula for the series is a(n) = 2^n + 3a(n-1), indicating that each term is derived from 2 raised to the power of n plus three times the previous term. This allows for straightforward calculation of any term by substituting the desired value of n. For example, the 5th term is calculated as a(5) = 80. This formula simplifies the process of solving the series. Further questions about the series are welcomed for clarification.
Poweranimals
Messages
68
Reaction score
0
Do you guys know what the recursion formula for this series?

http://ourworld.cs.com/SuperSamuraiStar/math.bmp
 
Last edited by a moderator:
Physics news on Phys.org
If

a_n = r^n

then

a_{n+1} = r^{n+1} = r \times r^n = r a_n

but you knew that! :-)
 


Hi there,

Thank you for sharing the link to the series. The recursion formula for this particular series is:

a(n) = 2^n + 3a(n-1)

This means that each term in the series is equal to 2 to the power of n, plus 3 times the previous term in the series.

Using this formula, you can easily calculate any term in the series by plugging in the value for n. For example, to find the 5th term in the series, you would plug in n=5:

a(5) = 2^5 + 3a(4)
a(5) = 32 + 3(2^4)
a(5) = 32 + 3(16)
a(5) = 32 + 48
a(5) = 80

So the 5th term in the series is 80.

I hope this helps and makes solving the series easier for you. Let me know if you have any other questions.

 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Replies
4
Views
3K
Replies
2
Views
3K
Replies
3
Views
2K
Replies
2
Views
2K
Replies
2
Views
2K
Back
Top