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September 30th, 2017, 03:44 PM  #1 
Newbie Joined: Jun 2017 From: Lima, Peru Posts: 15 Thanks: 0 Math Focus: Calculus  How to determine the nth term in this series?
I have this series, but it does not seem to follow a constant difference, or at least is not implicit $\displaystyle 1,0,5,14,27,...$ How can I determine the nth term, let's say the 16th term and the sum up to that term? Is there an easy way without resorting to induction? Last edited by skipjack; September 30th, 2017 at 08:50 PM. 
September 30th, 2017, 10:24 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,432 Thanks: 1462 
The nth term is 2nÂ²  5n + 2. (This is the simplest formula I could spot.) Hence the sum of the first n terms is (4nÂ³  9nÂ²  n)/6. The first 16 terms are 1, 0, 5, 14, 27, 44, 65, 90, 119, 152, 189, 230, 275, 324, 377, 434. Successive differences are 1, 5, 9, 13, 17, 21, 25, 29, . . . Successive second differences are 4, 4, 4, 4, 4, 4, 4, . . . 
September 30th, 2017, 11:33 PM  #3  
Newbie Joined: Jun 2017 From: Lima, Peru Posts: 15 Thanks: 0 Math Focus: Calculus  Quote:
The second part which involves the sum of terms in the sequence seems kind of logical as there exists these useful formulas (which can be obtained from any book). $\displaystyle \sum_{k=1}^{n}k^{3}=\left (\frac{n(n+1)}{2} \right )^{2}$ $\displaystyle \sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6} $ Last edited by skipjack; October 1st, 2017 at 12:06 AM.  
October 1st, 2017, 12:14 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,432 Thanks: 1462 
As the second differences are all 4, one can find a quadratic formula of the form 4nÂ²/2 + bn + c for the nth term. It's now easy to determine the values of b and c.

October 1st, 2017, 12:20 AM  #5 
Senior Member Joined: Sep 2015 From: USA Posts: 1,695 Thanks: 861 
I can't speak for how skipjack came up with the recursion, but you can always try to solve for the coefficients of a polynomial. Suppose you have $N$ points let $p(n)=\displaystyle \sum_{k=0}^{N1}~c_k n^k$ and you end up with a system of $N$ equations when you plug in values, e.g. in this case $1 = p(1),~0=p(2), 5=p(3), 14=p(4), 27=p(5)$ Solving this, you find $c_0 = 2,~c_1=5,~c_2=2,~c_3=0,~c_4=0$ Last edited by skipjack; October 1st, 2017 at 12:27 AM. 
October 3rd, 2017, 07:39 PM  #6  
Newbie Joined: Jun 2017 From: Lima, Peru Posts: 15 Thanks: 0 Math Focus: Calculus  Quote:
Can you show the example of how to use the sum to solve for, let's say, $c_1$? Sorry if I do ask it this way, but it is not too obvious for the casual learner. Last edited by skipjack; October 4th, 2017 at 12:56 PM.  
October 4th, 2017, 05:21 AM  #7 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,922 Thanks: 785 
Of course, a "sequence" doesn't have to follow any simple rule at all. When I was a freshman in college, my math professor gave an example of the sequence "15, 16, 17, 18, 31, 32, 33, 53, 54". Those were the numbers of the subway stations on his way home after work. His particular train shunted off one "line" to another twice. The best you can do with a problem like this is look for the "simplest" formula. And, of course, what is "simplest" may depend upon the person setting the problem. Last edited by skipjack; October 4th, 2017 at 12:58 PM. 
October 4th, 2017, 01:20 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 18,432 Thanks: 1462 
Using the formula 4nÂ²/2 + bn + c, putting n = 1 leads to 1 = 2 + b + c, and putting n = 2 leads to 0 = 8 + 2b + c. Subtracting the first equation from the second gives 1 = 6 + b, so b = 5. 
October 4th, 2017, 02:11 PM  #9 
Math Team Joined: Jul 2011 From: Texas Posts: 2,692 Thanks: 1351 
$\{1,0,5,14,27,...\}$ $\{1 \cdot 1, \, 0\cdot 3, \, 1 \cdot 5, \, 2 \cdot 7, \, 3\cdot 9, \, ... , \, (n2)(2n1), \, ... \}$ 
October 4th, 2017, 05:05 PM  #10 
Global Moderator Joined: Dec 2006 Posts: 18,432 Thanks: 1462 
I originally noticed that the nth term is $T_{2\text{n}3}$  1 â‰¡ (2n  3)(2n  2)/2  1 â‰¡ 2nÂ²  5n + 2.


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