An Action Research on Problem in finding the nth term of a sequence
By : Krishna Singh Pela
Instructor of Mathematics
Educational Training Center,"B"
Dhangadhi, Kailali, Nepal
Introduction
There are many types of sequences and series i.e. arithmetic sequence, geometric sequence where there is a certain pattern and the difference of two consecutive terms or the ratio of two consecutive terms is always equal or constant. In such sequences, we can easily find the nth term or the summation of the series by using the formulae. But there are some complicated sequences where there is neither the first difference nor the first ratio is constant. And if we try to find the nth term of such sequences, we can't get it easily. There are two options :
Use inspection/induction method and try until the result is found
Use some special formula and find it.
The problem is to find the general term of a given sequence or we can say it the nth term of the sequence.
Objective of the study
The main objective this study is to call the demand of trainee teachers of TPD (2070), of Kailali district and to make them able to teach this topic in their classroom. This is a general type of problem so the objective of this study is also focused towards preparing sufficient reference materials on this topic so that I will feel easy in my future training classes.
It's a textbook based problem of high school mathematics(optional math.)
The high school teachers have been facing a great problem on this topic. When they fill need proposal form for TPD, they put this problem each time.
There is no sufficient reference given in the textbook and the method is based on induction.
Data collection
List of teachers who put the demand| S.N. | Name of Teachers | Name of School |
| 1 | Puskar Raj Bhatta | Saraswati HSS, Geta |
| 2 | Govind Prasad Bhatta | Sharada HSS, Baiyabehedi |
| 3 | Bikalpa Chaudhari | Laxmi HSS, Udasipur |
| 4 | Asharam Chaudhari | Mahendra Shishu HSS, Dharmapur |
| 5 | Rajendra Pokharel | Khadgasmriti HSS, Tikapur |
| 6 | Jang Bahadur Bam | Bhanu ma.vi. Pathari |
| 7 | Harendraraj Awasthi | Janata Ra.ma.vi. Kailali gaun |
| 8 | Gyanendraraj Joshi | Trinagar HSS, Dhangadhi |
| 9 | Uddhav Dev Bhatt | Jankalyan HSS, Jkugeda |
| 10 | Janak Prasad Bhatt | Phulbari HSS, Phulbari |
| 11 | Karna Bahadur Kadayat | Gwasi HSS, Malakheti |
| 12 | Gorakh Nath | Shaileswori HSS, Badeha |
| 13 | Radha Bisht | Saraswati HSS, Pratappur |
| 14 | Raj Bahadur Kadayat | Siddsha Baba Ma.vi. Geti |
| 15 | Ganesh Datt Joshi | Panchoday HSS, Dhangadhi |
| 16 | Devendra Jha | Trinagar HSS, Dhangadhi |
| 17 | Raj Kamal Bhandari | Dhangadhi HSS, Dhangadhi |
| 18 | Prabhuram Chaudhari | Dipendra HSS, Hasuliya |
All these teachers were participants in the TPD training of 2069/70 at ETC 'B' , Dhangadhi, Kailali and all of them filled the demand proposal form before coming to this training. Some of these teachers have put their demand on the same topic i.e. nth term of a sequence.
i) Study the structure/pattern of the various sequences
ii) Limitations of the formula for nth term of AS and GS
iii) Consult reference materials from various books and websites
iv) Organize the matter and use it in the training class.
i) Study the structure/pattern of the various sequences Case I :
Let us have a sequence 1, 3, 5, 7, …, …, …
Here d = 2, so nth term will be multiple of 2 and to find first term = 1 from 2n (for n = 1), we should subtract 1.
Thus nth term will be 2n-1.
We can also check as follows :
t1 = 1 = 2.1-1 , t2 = 3= 2.2-1
t3 = 5 = 2.3 -1 , t4 = 7 = 2.4 -1
…. … … , tn = 2.n-1
Case II : Let us have another sequence
1, 2, 4, 8, 16, … , …,
t1 = 1.21-1= 1.20 = 1.1 = 1,
t2= 1.22-1= 1.21= 1.2= 2
similarly tn = 1.2n-1 = 2n-1
Direct formulae for arithmetic sequences is tn = a + (n-1)d and for geometric sequcences is tn = arn-1, but the important thing to remember is that we use the induction method rather than the direct formula. However, the formula helps us to think of the certain pattern and it saves our time and effort and we can easily guide the students to go in a appropriate way.
ii) Limitations of the formula for nth term of AS and GS
Let us have another type of sequence
10, 12, 16, 22, 30, …, …
Note that this sequence is neither in AS nor in GS and the formula which we used above do not work here. This is the limitation of the formulae. At this stage again either we have to use induction method or search for a new formula.
iii) Findings from reference materials, various books and websites
Finally I made up mind to go under the shelter of internet. I got the following formulae, tips and conclusions :
1. Common difference formula/when the first differences are constant (i.e. the sequence is arithmetic)
Imagine the sequence 2, 4, 6, 8, 10, …
The formula used here is a + (n-1)d
2. For geometric sequence tn = arn-1
3. Changing difference formula
Here,specially When the II differences are constant,
tn = a +(n – 1)d + 0.5(n -1)(n -2) C
Where d = the first difference =t2- t1,
C = c. d. of the sequence of first differences
Here d = 5 -3 =2, c = 2
Nth term = 3 + (n – 1)x2 + 0.5(n – 1)(n – 2)x2
= 3 + 2n – 2 + n2 - 3n + 2 = n2 - n + 3
Implementation
To use this information in the training classroom, I porepared some worksheets, wrote t he related formulae on it and divided the students in 6 groups of 3 participants in each.The group division was done on the following basis :
Group serial no. (above table of name list)
A (1, 3, 5)
B (7, 9, 11)
C (13, 15, 17)
D (2, 4, 6)
E (8, 10, 12)
F (14, 16, 18)
The worksheet was same for each group and was as follows :
Find the nth term of the following sequences :
i) 8, 3, -2, -7, -12, …, …,
ii) 3, 6, 12, 24, 48, …, …
iii) 4, 9, 16, 25, 36, …, …
iv) 5, 10, 17, 26, 37, 50, …, …
v) 6, 11, 18, 27, 38, 51, …, …
vi) 3, 8, 15, 24, 35, 48, …, …
vii) -4, 9, -16, 25, -36, 49, …, …
viii) 4, -9, 16, -25, 36, …, …
ix) 2/4, 5/9, 8/16, 11/25, …, …
x) -5, 7/2, -9/3, 11/4, -13/5, …, …
xi) 1, 1/2 , 1/3, 1/4, 1/5, …, …
xii) 1/2, 2/3, 3/4, 4/5, …, …
xiii) -3, -1, -1/3, 0, 1/5, …, …
xiv) -1, 2, -3, 4, -5, …, …
xv) 3, -2, 5/3, -3/2, 7/5, …, …
xvi) 1, 1/2, 3, 1/4, 5, …, …
xvii) 1/2, -2/6, 3/18, -4/54, 5/162, …, …
xviii) 5/6, 7/12, 9/24, 11/ 48, 13/96, …, …
xix) 1, 5, 14, 30, …, …
xx) 1, 9, 36, 100, …, …
Conclusion
After the performing the group work, an individual evaluation was also conducted and each group achieved the competency in finding the general term of the given sequence. They also made the following conclusions :
1. Check if the numbers are square numbers like 4, 9, 16, 25, 36, …, write them as 22, 32, 42, 52, … Note that bases are in A.S., where d = 1 and exponent is a constant, so bn= 2+ (n-1)x1= n+1
and so tn = (n+1)2
2) If the second differences are equal then the nth term begins with n2 i.e. for 1, 4, 9, 16, …,
tn = n2
3) If third differences are equal then nth term begins with n3 i.e. for 1, 8, 27, 64, …,
tn = n3
4) If the terms of sequence are numbers next to perfect squares i.e.
5, 10, 17, 26, 37, …
= 22+1, 32+1, 42+1, …
Now find nth term as in above example and add 1.
Thus tn = (n+1)2 +1
5) Similarly for 6, 11, 18, 27, 38, …
= 22+2, 32+2, 42+2, …
tn = (n+1)2 +2
6) For the sequence 3, 8, 15, 24, … =22-1, 32-1, 42-1, …
tn = (n+1)2 -1
7) Check if the terms of the sequence are fractional and it is not an arithmetic or geometric sequence, then calculate the nth term of the numerator and denominator separately. i.e. for 2/4, 5/9, 11/25, …,
here tn = (3n-1)/(n+1)2
8)Note that there is no easy way of working out the nth term of a sequence other than to try different possibilities.
Unsolved topic that will be studied in the next cycle
Although the trainee teachers got the skill to find the general term for the sequence which has first and second differences constant, but the above explanation doesn't solve the problem if the third or fourth or further differences are equal/constant.
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