## Friday, June 12, 2015

### [H2_20150512SSSTRR] Finding a Recurrence Relation for Terms in a Series

Question

Introduction
This question pertains to the relationship between the partial sums of a series and its terms.  I am not sure if all the junior colleges teach this explicitly, but students are expected to know or be able to observe this relationship.  Let us follow our nose and focus on the first part first.

Reminders
For the series  u1 + u2 + ¼ + un–1 + un + ¼  ,  the nth partial sum
Sn = u1 + u2 + ¼ + un–1 + un
Sn–1 = u1 + u2 + ¼ + un–1
Taking the difference, we see that
un = SnSn–1
Innocuous looking, this is actually a very powerful formula.  It is applicable to all sequences and series (not only for arithmetic and geometric series).  That means this formula can always be used!
Another thing to note is that sequences  un  and partial sums  Sn  (which are themselves another sequence) behave like functions.  [In advanced mathematics, they are in fact defined as functions with domain as the positive integers.]  What this means is that  Sn-1  has the same formula as  Sn  except that  n  is replaced with  (n – 1).

Solution

Checking
Actually, the question setter forgot that the formula works for  n > 1.
OK, let us check whether the formula really works.  We know that  u1 = 3.  Let us tabulate and compare the recursive formula with the explicit formula.  You can do this on a piece of rough paper.

 n recursive un = f(un–1) explicit un = 3´2n–1 1 u1 = 3 3´21–1 =  3 2 u2 = 2´  3   = 6 3´22–1 =  6 3 u3 = 2´  6  = 12 3´23–1 = 12 4 u4 = 2´12 = 24 3´24–1 = 24 5 u5 = 2´24 = 48 3´25–1 = 48

Challenge
What if the question wanted a recurrence relation for  Sn?

H04. Look for pattern(s)
H13* Use Equation / write a Mathematical Sentence

Suitable Levels
GCE ‘A’ Levels, H2 Mathematics
International Baccalaureate Mathematics
* other syllabuses that involve sequences and series