Discover the nth Term and Sum of a Series: 12, 23, 60, 169, 494

  • Thread starter Thread starter hav0c
  • Start date Start date
  • Tags Tags
    Series
AI Thread Summary
The discussion focuses on finding the nth term and the sum of the series 12, 23, 60, 169, 494. It is noted that each term can be expressed as three times the previous term minus a constant, leading to a proposed formula for the nth term. However, the user struggles with deriving a complete formula and ends up with an unresolved tn+1 term. Suggestions include solving the homogeneous recurrence relation and adding a linear term with unknown constants to refine the approach. The conversation emphasizes the need for clarity in deriving the nth term and sum of the series.
hav0c
Messages
58
Reaction score
0

Homework Statement



Find the nth term and sum to n terms of the series 12,23,60,169,494..

Homework Equations





The Attempt at a Solution


well each term is the triple of the previous term minus some constant.
23=10*3-7
60=23*3-9
..
..
This way i am able to write the nth term as 3(tn-1)-3-2n
and hence proceed to find a telescopic sum by rearrangement and adding.
but this method leaves me with a tn+1 which i am unable to write
So how shud i proceed
 
Physics news on Phys.org
hav0c said:
Find the nth term and sum to n terms of the series 12,23,60,169,494..
I guess you meant 10,23, ...
well each term is the triple of the previous term minus some constant.
23=10*3-7
60=23*3-9
..
..
This way i am able to write the nth term as 3(tn-1)-3-2n
and hence proceed to find a telescopic sum by rearrangement and adding.
but this method leaves me with a tn+1 which i am unable to write
So how shud i proceed
Please post your working as far as you got.
Can you solve the homogeneous recurrence relation?
To account for the extra -3-2n, just try adding a linear term with a couple of unknown constants. See if you can determine what those constants would need to be.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

Similar threads

Back
Top