Page 31 - PRE-U STPM MATHEMATICS (T) TERM 1

Mathematics Term 1 STPM Chapter 2 Sequences and Series

Example 28

Find the r term, u , of the series 2 · 1! + 5 · 2! + 10 · 3! + 17 · 4! + … + (n + 1) · n!
2
th
r
By expressing u as the difference of two functions of r, find the sum of the above series.
r
Solution: For the series
2 · 1! + 5 · 2! + 10 · 3! + 17 · 4! + … + (n + 1) · n!,
2
th
2
the r term, u = (r + 1) · r!
r
= [r(r + 1) – (r – 1)] · r!
2 = r(r + 1)r! – (r – 1) · r!
= r(r + 1)! – (r – 1)r! (r + 1)r ! = (r + 1)!

Let f(r) = r(r + 1)! and f(r – 1) = (r – 1)r!
Then u = f(r) – f(r – 1)
r
n n
and ∑ u = ∑ [f(r) – f(r – 1)]
r = 1 r r = 1
= f(n) – f(0)
= n(n + 1)! – 0
n
2
∴ ∑ (r + 1) r! = n(n + 1)!
r = 1

Exercise 2.5

1. Evaluate
4 12 8
(a) ∑ (r + 3r) (b) ∑ (100 – r ) (c) ∑ (4r + 5)
2
3
r = 1 r = 10 r = 3
5 90 8 rπ 5
r
(d) ∑ (e) ∑ sin (f) ∑ (–1) (1 + 2 r + 1 )
r = 1 r r = 1 3 r = 0
n n n n
2
3
2. Using the results for ∑ r, ∑ r and ∑ r , find the value of ∑ u for each of the following cases.
r = 1 r = 1 r = 1 r = 1 r
2
3
(a) u = 2r – r + 1 (b) u = r (r + 2)
r
r
(c) u = (r + 2)(r + 3)(2r – 1) (d) u = (r + 1)(r + 3)(r + 5)
r
r
3. For each of the following series, write down its r term. Hence, find the sum up to the n term.
th
th
(a) 1 · 4 + 2 · 7 + 3 · 10 + … (b) 1 · 5 + 2 · 6 + 3 · 7 + …
2
2
2
(c) 1 · 3 + 2 · 4 + 3 · 5 + … (d) 1 · 4 · 7 + 4 · 7 · 10 + 7 · 10 · 13 + …
(e) 1 + 1 + 1 + … (f) 1 · 2 · 3 + 2 · 3 · 4 + 3 · 4 · 5 + …
1 · 2 · 3 2 · 3 · 4 3 · 4 · 5
2
2
2
2
4. Given that 1 + 2 + 3 + … + n = 1 n(n + 1)(2n + 1), find the sum of the series
6
2
2
2
2 + 4 + 6 + … + 50 2
Using the above result, find the value of
2
2
2
2
1 + 3 + 5 + … + 49 .
3
3
3
3
5. Given that 1 + 2 + 3 + … + n = 1 n (n + 1) , find the sum of the first twenty terms of the series
2
2
4
2 + 16 + 54 + 128 + 250 + …
118
02 STPM Math T T1.indd 118 3/28/18 4:21 PM