Page 27 - PRE-U STPM MATHEMATICS (T) TERM 1

Mathematics Term 1 STPM Chapter 2 Sequences and Series

(c) By using
n 1
∑ r = n (n + 1) 2
2
3
r = 1 4
and substituting n = 25,
25
3
2
∑ r = 1 (25) (26) 2
r = 1 2
= 105 625

2 Example 23

2
2
Find the sum of the series 1 + 3 + 5 + 7 + … to n terms.
2
2
2
2
2
2
Hence, find the value of 1 + 3 + 5 + … + 99 .
2
2
2
Deduce the value of 51 + 53 + … + 99 .
Solution: The sum of the squares of the odd integers is
2
2
2
2
1 + 3 + 5 + 7 + …
The r term is (2r – 1) 2
th
n
2
Hence, the sum of the series is ∑ (2r – 1) .
r = 1
n n
2
∑ (2r – 1) = ∑ (4r – 4r + 1)
2
r = 1 r = 1
n n n
2
= 4 ∑ r – 4 ∑ r + ∑ 1
r = 1 r = 1 r = 1
4
4
= 4 3 n (n + 1)(2n + 1) – 4 3 n (n + 1) + n
2
6
= n [2(n + 1)(2n + 1) – 6(n + 1) + 3]
3
2
= n (4n + 6n + 2 – 6n – 6 + 3)
3
2
= n (4n – 1)
3
50 th
2
2
2
2
1 + 3 + 5 + … + 99 = ∑ (2r – 1) 2 n term, 2n – 1 = 99
r = 1 n = 50
2
= 50 [4(50) – 1]
3
= 50 × 9999
3
= 166 650
2
2
2
2
2
2
2
2
2
51 + 53 + … + 99 = (1 + 3 + … + 99 ) – (1 + 3 + … + 49 )
50 25
= ∑ (2r – 1) – ∑ (2r – 1) 2
2
r = 1 r = 1
2
= 166 650 – 25 [4(25) – 1]
3
= 166 650 – 20 825
= 145 825
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02 STPM Math T T1.indd 114 3/28/18 4:21 PM