Page 30 - PRE-U STPM MATHEMATICS (T) TERM 1

Mathematics Term 1 STPM Chapter 2 Sequences and Series

n n
∑ 1 = ∑ 1 1 – 1 2
r = 1 r(r + 1) r = 1 r r + 1
n
= – ∑ 1 1 – 1 2
r = 1 r + 1 r
Let f(r) = 1 , f(r – 1) = 1
r + 1 r
n n
Thus ∑ 1 = – ∑ [f(r) – f(r – 1)]
r = 1 r(r + 1) r = 1
= –[f(n) – f(0)]
1
= – 1 n + 1 – 1 2 2

= 1 – 1
n + 1
= n
n + 1

Example 27

Show that 1 – 1 = – 2 .
(r + 1)(r + 2) r(r + 1) r(r + 1)(r + 2)

n
Using the method of differences, find ∑ 1 .
r = 1 r(r + 1)(r + 2)

Solution: 1 – 1 = r – (r + 2)
(r + 1)(r + 2) r(r + 1) r(r + 1)(r + 2)
= – 2
r(r + 1)(r + 2)

1
Thus 1 = – 3 1 – 1 4
r(r + 1)(r + 2) 2 (r + 1)(r + 2) r(r + 1)
1
1
Let u = r(r + 1)(r + 2) and f(r) = (r + 1)(r + 2) .
r
1
∴ u = – [f(r) – f(r – 1)].
r
2
n n
1
and ∑ u = – ∑ [f(r) – f(r – 1)]
r = 1 r 2 r = 1
1
= – [f(n) – f(0)]
2
1
= – 3 1 – 1 4
2 (n + 1)(n + 2) 2
n
3

∑ 1 = 1 1 – 1 4 .
r = 1 r(r + 1)(r + 2) 2 2 (n + 1)(n + 2)

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