Page 28 - PRE-U STPM MATHEMATICS (T) TERM 1

Mathematics Term 1 STPM Chapter 2 Sequences and Series

Example 24

th
Find the r term of the series
1 · 2 · 3 + 2 · 3 · 4 + 3 · 4 · 5 + …
n n
2
By using the results for ∑ r and ∑ r , find the sum of the first n terms of the above series.
r = 1 r = 1
Solution: The series is
1 · 2 · 3 + 2 · 3 · 4 + 3 · 4 · 5 + …
th
The r term is r(r + 1)(r + 2).
st
The sum of the 1 n terms of the series is
n n 2
2
3
∑ r(r + 1)(r + 2) = ∑ (r + 3r + 2r)
r = 1 r = 1
n n n
3
= ∑ r + 3 ∑ r + 2 ∑ r
2
r = 1 r = 1 r = 1
4
4
2
= n 2 (n + 1) + 3 3 n (n + 1)(2n + 1) + 2 3 n (n + 1)
4 6 2
= n (n + 1)[n(n + 1) + 2(2n + 1) + 4]
4
2
= n (n + 1)(n + 5n + 6)
4
= n (n + 1)(n + 2)(n + 3)
4
Summation of series using the method of differences
Consider the sum of the terms of a series such as
u + u + u + … + u n
3
2
1
th
Suppose that the r term, u , can be expressed in the form f(r) – f(r – 1), where f(r) is a function of r.
r
n n
Then ∑ u = ∑ [f(r) – f(r – 1)]
r = 1 r r = 1
= f(n) – f(n – 1)
+ f(n – 1) – f(n – 2)
+ f(n – 2) – f(n – 3)
+ … Substituting the values of
+ f(3) – f(2) r, term by term, with r = n,
n – 1, n – 2, …, 3, 2, 1
+ f(2) – f(1)
+ f(1) – f(0)
= f(n) – f(0)
th
This means that if the r term, u , of a series can be expressed as the difference of a function of r and a function
r
n
of (r – 1), then the sum of the series, ∑ u , can be determined.
r = 1 r
This method of finding the sum of a series is called the method of differences.
n
If u = f(r) – f(r – 1), then ∑ u = f(n) – f(0)
r
r
r = 1

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02 STPM Math T T1.indd 115 3/28/18 4:21 PM