Page 26 - PRE-U STPM MATHEMATICS (T) TERM 1

Mathematics Term 1 STPM Chapter 2 Sequences and Series

2
= 1 (n + 1)[2(n + 1) – 2 – 3n]
2
2
= 1 (n + 1)(2n + 4n + 2 – 2 – 3n)
2
2
= 1 (n + 1)(2n + n)
2
= 1 n(n + 1)(2n + 1)
2
n
2
Thus ∑ r = 1 n(n + 1)(2n + 1)
r = 1 6
Similarly, by using the identity
3
4
4
2
(r + 1) – r = 4r + 6r + 4r + 1, 2
n 1
we get ∑ r = n (n + 1) 2
3
2
r = 1 4
= 3 1 2 n(n + 1) 4 2
n 2
1
= ∑ r 2
r = 1
Sum of the first n positive integers is
n 1
∑ r = n(n + 1).
r = 1 2
Sum of the squares of the first n positive integers is
n
∑ r = 1 n(n + 1)(2n + 1)
2
r = 1 6
Sum of the cubes of the first n positive integers is
n 1
2
3
2
∑ r = n (n + 1) .
r = 1 4
Example 22
Evaluate
100 50 25
(a) ∑ r (b) ∑ r 2 (c) ∑ r 3
r = 1 r = 1 r = 1
Solution: (a) By using
n 1
∑ r = n(n + 1)
r = 1 2
and substituting n = 100,
100 1
∑ r = (100)(101)
r = 1 2
= 5050
(b) By using
n 1
2
∑ r = n(n + 1)(2n + 1)
r = 1 6
and substituting n = 50,
50 1
∑ r = (50)(51)(101)
2
r = 1 6
= 42 925

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02 STPM Math T T1.indd 113 3/28/18 4:21 PM