F6 Mathematics T
1
Revision Notes on Chapter 2 : Sequences & Series (Term 1) Name : ___________________ ______________________________ ___________
Date : __________________ __________________
2.1 : Sequences
1). Limit of a sequence = lim un n
2). If lim un exist, the sequence is a convergent sequence or otherwise it will be a divergent sequence. n
2.2 : Series (A) : Arithmetic Progression ( A. P. )
1). The n th. Term, un a ( n 1 ) d st 2). Sum of the 1 . n terms, Sn
n 2
( 2a ( n 1)d )
Sn
or
n 2
( a l)
Proof : i ). Sn a ( a d ) ( a 2d ) ... ( a ( n 1) d ) ------ (1) ii ). Sn ( a (n 1) d ) ( a ( n 2) d ) ( a ( n 3) d ) .. ... a ------ (2) iii). (1) + (2) : 2 Sn ( 2a ( n 1) d ) (2 ( 2a (n 1) d ) (2 ( 2a (n 1) d ) .. ... (2 (2a (n 1) d ) 2 Sn n ( 2a (n 1) 1) d ) n Sn ( 2a (n 1) d ) 2 3). If A. P. : …, p, m, q, … , then The Arithmetic mean of p p and q
p q 2
m
(B) : Geometric Progression ( G. P. )
1). The n th. Term, un a r n 1 st
2). Sum of the 1 . n terms,
Sn a (
rn 1 r 1
)
Sn a (
or
( For r > 1 )
1 r
)
( For r < 1 )
Proof : i ). Sn a ar ar 2 ... ar n 1
------ (1)
ii ). rSn ar ar 2 ar 3 ... ar n ------ (2) iii). (1) - (2) :
Sn ( 1 r ) a ar n Sn a (
1 r n
1 r n 1 r
3). If G. P. : …, p, m, q, … , then The Geometric mean of p p and q pq m
)
4). A geometric series is convergent when
1 and the sum to infinity, S
r
(C) : Summation (Sigma) Notation n
1).
u
u1 u2 u3 ... un
r
r 1 n
2).
n
ku
k ur
r
r 1
r 1
n
3).
k
n k
r 1 n
4).
u
r
vr
n
u v
r 1
5).
n
r
r 1
r
r 1
r The term ( 1) will give the alternating sign of a series.
st
(D) : Sum of the 1 . n term (Summation) of other Series (1) : Natural Number Series
i ). Natural number are 1, 2, 3, 4, 5, … n
ii ).
n
r 2 n 1 r 1
n
Proof:
a).
r 1 2 3 ... (n 1) n
------ (1)
r 1 n
b).
r n (n 1) (n 2) ... 2 1
------ (2)
r 1 n
c). (1) + (2) :
2
r n(n 1)
n
r 1 n
iii).
r
2
n
r 2 n 1
r 1
n
n 1 2n 1
r 1
Proof:
6
3
a). r 1 r 3 3r 2 3r 1 3
b). r 1 r 3 3r 2 3r 1 n
c).
r 1
3
r 1
d). n 1 13 3
r 1 n
r
e). 3
2
r 1
2
f). 3
n
g).
r r 1
2
r 1
r 1
3
2
3
3
n 1 (n 1) n(n 1) 2
r2
r 1
n
n(n 1) n
r 1 n
n
2
n
r
3
n
r 3 r 3 r 1 3
1 2 n
(n 1)(2n 2 4n 2 2 3n )
n 1 2n 1 6
a 1 r
2
n
iv).
r
3
r 1
n2 4
(n 1)2 3 n
Proof : almost the same as for
r r 1
2
n
n 1 2n 1 6
v ). Steps to find sum to n terms of the natural number series: n
a) Write the given u1 u2 u3 u 4 ... series in the form of
u
r
where ur can be found by using the
r 1
nth. term formulae of the A.P. or G.P. or the combination of both, etc. b) Eliminate the summation symbol by using formulae in (D)(ii), (iii) & (iv). c) Factorize & simplify the sum to n terms.
(2) : Other Series which can be simplified by “Method of Differences” n
i ).
ii ).
iii).
iv).
n
u ( f (r 1) f (r )) f (n 1) f (1) r
r 1
r 1
n
n
u ( f (r ) f (r 1)) f (1) f (n 1) r
r 1
r 1
n
n
u ( f ( r ) f (r 1)) f (n) f (0) r
r 1
r 1
n
n
u ( f ( r 1) f (r )) f (0) f (n) r
r 1
r 1
v ). To show that a series is convergent, (taking case (2)(i) for example) when
n ,
f (n 1) f (1) (in simplified form)
a (find a) (if a is a constant, then the series is convergent)
S a vi). Other than formulae (2)(i) to (iv) for “Method of Differences”, sum to n term of a certain series can also be simplified by : st a). Expanding the summation by listing the sum of the 1 . & last few terms of the series. b). Drop the terms which can be eliminated by themselves and simplify.
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2.3 : Binomial Expansions (Expansions of the form (
a + b )
n
)
n
(A). Expand ( a + b ) , where n is a +ve integer (Finite Series)
1). The binomial coefficients for a binomial expansion ( a b )n with small n value can be obtained from the Pascal’s Triangle :
1 1 1 1
1
1 2
3 4
1 3
6
1 4
1
2). When n Z ,
(a b ) n a n
n 1!
a n 1b
n (n 1) 2!
a n 2b 2 ...
n (n 1)(n 2)...(n r 1) r !
a n rb r ... nab n 1 b n
n n n n n 1 n a n a n 1b a n 2b 2 ... a n rb r ... ab b 1 2 r n 1 Note : the above expansion has a total of n+1 terms.
n n! 3). The coefficient of ( r 1) th. term r r !( n r )! 4).
n n r = n r
5).
n n n 1 r + r 1 = r 1
6). Specific term of an expansion,
n
C r
n ur 1 a n r b r r
Must know: 9
3 i ). For p q , find the term with p .(Hint: 9 r 3, r 6, find u 61 )
ii ). For ( a b )8 , find the middle term .(Hint: total term = 8+1 =9,middle term= u5
u4 1 )
12
3 1 iii). For 8 x , find the term independent of x. (Hint: 3(12 r ) ( 1)r 0 , r 9 ,find u9 1 ) 2 x 4 4 iv). Find the coefficient of x 3 in the expansion of 4 3 x 3 2 3 x . (Hint: in the expansion of 2 3 x ,
find 1 term independent of x and 1 term with x 3 only. Then, multiply the term independent of x with 3 x 3 and the other term with 4 & get the total.)
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(B). Expand ( 1 +
x )
n
, where n is a +ve integer (Finite Series)
1). When n Z ,
(1 x ) n 1
n 1!
x
n (n 1) 2!
x 2 ...
n (n 1)(n 2)...(n r 1) r !
x r ... nx n 1 x n
n n n n n 1 n 1 x x 2 ... x r ... x x 1 2 r n 1 Note : the above expansion has a total of n+1 terms. 2). b a b a 1 a n
n
n
= a n 1 x
n
, where x
b a
Note : the technique above is used when finding an approximation with n ve integer .
(C). Expand ( 1 +
x )
n
, where n is NOT a +ve integer (Infinite Series)
1). If n Z , (e.g. n 2, (1 x ) n 1
n 1!
x
2
etc.) 3 n (n 1) 2!
x2
n (n 1)(n 2) 3! n
i ). This expansion is only valid for 1 x when
x 3 ...
n (n 1)(n 2)...(n r 1) r !
x r ...
n
x 1 and not valid for a b .
ii ). This expansion is used when finding approximation with n not a +ve integer. n
b 2). To use this expansion, a b a 1 , where a n
b
n
a
<1.
Must know: i ). Using Binomial Theorem to find
3
8.72 correct to 4 d.p..(Hint:
3
3
stop the expansion until the term with 0.09 for 4 d.p. accuracy.) ii). if
3 x
< 1 , find the range of x.
1
3).
1 x 1 x x 2 x3 ... ( 1)r x r ...
4).
1 x 1 x x 2 x3 ... xr ...
1
,
,
1 3
8.72 8(1 0.09) 2(1 0.09) ;
x 1
x 1
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