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 u_{1} + u_{2}
+ ¼ + u_{n}_{–1} +
u_{n} + ¼ , the n^{th}
partial sum
S_{n} = u_{1} + u_{2}
+ ¼ + u_{n}_{–1} +
u_{n}
S_{n}_{–1} = u_{1} + u_{2}
+ ¼ + u_{n}_{–1}
Taking the difference, we see that
u_{n} = S_{n}
– S_{n}_{–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 u_{n} and partial sums S_{n}
(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 S_{n}_{1} has the same formula as S_{n}
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 u_{1}
= 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
u_{n} = f(u_{n}_{–1})

explicit
u_{n} = 3´2^{n}^{–1
}

1

u_{1} = 3

3´2^{1–1} = 3

2

u_{2} = 2´ 3 = 6

3´2^{2–1} = 6

3

u_{3} = 2´
6 = 12

3´2^{3–1} = 12

4

u_{4} = 2´12 = 24

3´2^{4–1} = 24

5

u_{5} = 2´24 = 48

3´2^{5–1} = 48

Challenge
What if the
question wanted a recurrence relation for
S_{n}?
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
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