Page 35 - PRE-U STPM MATHEMATICS (T) TERM 1

Mathematics Term 1 STPM Chapter 2 Sequences and Series

= n! 3 r + 1 + n – r 4
(n – r – 1)! r! (n – r)(r + 1)
= (n + 1) n!
(n – r)(n – r – 1)! (r + 1)r!
= (n + 1)!
(n – r)! (r + 1)!
(n + 1)!
1 n + 1 2 = [(n + 1) – (r + 1)]! (r + 1)!
r + 1
= (n + 1)!
(n – r)! (r + 1)!
n
n
2 Hence, 1 2 1 r + 1 2 1 n + 1 2
=
+
r
r + 1
n
Both the above results (a) and (b) are very useful in evaluating 1 2 , especially when the difference between
r
n and r is small.
For any positive integers n and r, n  r,
n
n
(a) 1 2 1 n – r 2
=
r
n
n
=
(b) 1 2 1 r + 1 2 1 n + 1 2
+
r + 1
r
Example 33
Evaluate
7
18
15
15
12
12
+
(a) 1 2 (b) 1 2 (c) 1 2 1 2 (d) 1 2 1 2
+
3
4
4
15
5
7
7
7!
Solution: (a) 1 2 = (7 – 3)! 3!
3
= 7!
4! 3!
= 7 × 6 × 5 × 4!
4! × 3 × 2 × 1
= 35
18
18!
(b) 1 2 = (18 – 15)! 15!
15
= 18 × 17 × 16 × 15!
3! × 15!
= 18 × 17 × 16
3 × 2 × 1
= 816
15
15
16
=
(c) 1 2 1 2 1 2 By using the result n + 1 2
+
2 1
1 2 1
n
n
4
5
5
=
+
= 16! r r + 1 r + 1
11! 5!
= 16 × 15 × 14 × 13 × 12 × 11!
11! × 5 × 4 × 3 × 2 × 1
= 4368
122
02 STPM Math T T1.indd 122 3/28/18 4:21 PM