Page 29 - PRE-U STPM MATHEMATICS (T) TERM 1

Mathematics Term 1 STPM Chapter 2 Sequences and Series

Example 25

Show that
r(r + 1)(r + 2)(r +3) – (r – 1)r(r + 1)(r + 2) = 4r(r + 1)(r + 2).
n
Hence, find ∑ r(r + 1)(r + 2).
r = 1
Solution: r(r + 1)(r + 2)(r + 3) – (r – 1)r(r + 1)(r + 2)
= r(r + 1)(r + 2)[(r + 3) – (r – 1)]
= 4r(r + 1)(r + 2)
2 Let f(r) = r(r + 1)(r + 2)(r + 3)
Then f(r – 1) = (r – 1)r(r + 1)(r + 2)
f(r) – f(r – 1) = r(r + 1)(r + 2)(r + 3) – (r – 1)r(r + 1)(r + 2)
= 4r(r + 1)(r + 2)
n
For the series ∑ r(r + 1)(r + 2), u = r(r + 1)(r + 2).
r = 1 r
f(r) – f(r – 1) = 4u r
or u = 1 [f(r) – f(r – 1)]
4
r
n n
∴ ∑ u = 1 ∑ [f(r) – f(r – 1)]
r = 1 r 4 r = 1
= 1 [f(n) – f(0)]
4
= 1 [n(n + 1)(n + 2)(n + 3) – 0]
4
n
Thus, ∑ r(r + 1)(r + 2) = 1 n(n + 1)(n + 2)(n + 3).
r = 1 4

Note: Compare the method of differences shown in Example 25 with the method shown in Example 24.
The method of differences can also be used if an expression can be expressed in partial fractions, as shown in
the following example.

Example 26

n
Express 1 in partial fractions. Hence, find ∑ 1 .
r(r + 1) r = 1 r(r + 1)
Solution: Let 1 ≡ A + B
r(r + 1) r r + 1
∴ 1 ≡ A(r + 1) + Br
Let r = 0 : 1 = A
Let r = –1 : 1 = –B
B = –1
Thus 1 = 1 – 1
r(r + 1) r r + 1

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