Page 45 - PRE-U STPM MATHEMATICS (T) TERM 1

Mathematics Term 1 STPM Chapter 2 Sequences and Series
th
4. For a geometric progression with first term a and common ratio r, the n term is u = ar n – 1 .
n
Sum of the first n terms is n
S = a(r – 1) , for r > 1
r – 1
n
n
and S = a(1 – r ) , for r < 1
1 – r
n
lim
n
For |r| , 1 and large values of n such that x → ∞ r → 0, the sum (to infinity) is
a
S = 1 – r
∞
n
n!
5. 1 2 = (n – r)! r! , n, r  Z , 0  r  n.
+
r
2 = n(n – 1)(n – 2) … (n – r + 1)
r!
n
n
1 2 = 1 n – r 2
r
n
n
1 2 1 r + 1 2 1 n + 1 2
=
+
r + 1
r
6. The binomial theorem
n
n
n
n
2
r
n
(1 + x) = 1 + 1 2 x + 1 2 x + … + 1 2 x + … + 1 n – 1 2 x n – 1 + x , for n  Z .
+
n
r
1
2
(1 + x) = 1 + nx + n(n – 1) x + … + n(n – 1) … (n – r + 1) x + … , for n  Q.
r
n
2
2! r!
STPM PRACTICE 2
1. Write down the n term of the sequence 0, 2, 6, 12, 20, …. Find n, if the n term is 210.
th
th
th
2. Find the first three terms and the n term of the series with the sum to n terms, S , given by
n
1
(a) S = 4n – n (b) S = (n + 1)!
n
n
3. Find the value of
30 8 n n
(a) ∑ (r + 2) (b) ∑ 2 r (c) ∑ 2r (d) ∑ r(r!)
r = 1 r = 1 r = 1 r = 1
n
n
4. Show that ∑ (r + 1)2 r – 1 = n2 .
r = 1
n
1
2
5. (a) Prove that ∑ r = n(n + 1)(2n + 1).
r = 1 6
(b) If S = 1(n) + 2(n – 1) + 3(n – 2) + … + r(n + 1 – r) + … + n(1)
and T = 1(n – 1) + 2(n – 2) + 3(n – 3) + … + r(n – r) + … + (n – 1)(1),
where n is a positive integer, show that
S + T = 1 n(n + 1)(2n + 1)
6
4
1
6. If u = (2n + 1)(2n + 3) , show that u n – 1 – u = (2n – 1)(2n + 1)(2n + 3) .
n
n
Hence, find the sum to n terms of the series
1 + 1 + … + 1 + …
1 · 3 · 5 3 · 5 · 7 (2r – 1)(2r + 1)(2r + 3)
132
02 STPM Math T T1.indd 132 3/28/18 4:21 PM