SEQUENCES AND SERIES
See the following sets of numbers, do they look cumbersome?
A A 1,5,9,13,17,21,25,29, ' B A1 ,2 ,3 , 4 ,5 ,6 ,7 ,8 , ' 1 f ff 1 f 1 f 1 f 1 f 1 f ff 1f ff ff ff ff ff ' ff ff ff ff ff ff ff ff ff ff ff ff ff ff f f f f f f
2 2 2 2 2 2 2 2
CA
, , , , , , , 10 11 12 13 14 15 16
But look again closely. Do you see a specific order or pattern in each set? Isn’t it easy to predict any term (say the 20th) in each of the above? We note that in A we add 4 each time to get the next number, in B we take the square of the next integer and in C we take the reciprocals of the next integer. Thus the 20th term in A is 77, in B is 1f 2 ff ff . 20 and in C is ff These are called sequences, where the subsequent terms are decided 39 by a given relation (which is expressed as a function f(n) ).
SEQUENCE
A sequence is a function f whose domain is the set of all natural numbers (infinite sequence) or some subset of natural numbers from 1 up to some larger number (finite sequence). The values f(1), f(2), f(3). . . . are the terms of the sequence.
Let the terms of a sequence be given by the function by f n = n 2 + 1 where n is a natural number A Here f 1 = 1 + 1 = 2,
` a ` a
2 2
f 2 = 2 + 1 = 5,
` a ` a
2 2
` a
f 3 = 3 + 1 = 10,
` a
2
f 4 = 4 + 1 = 17, f 5 = 1 + 1 = 26, ……A The sequence becomes 2, 5, 10, 17, 26, ………………A
1, 5, 9, 13, 17 is a finite sequence with 5 terms. 1, 5, 9, 13, 17, 21,……… is an infinite sequence. A sequence is represented by its range. f n = an is used to denote the range elements of the function. a1 , a2 , a3 , ……an , …A are the first term, second term, third term,….nth term,…. respectively of the sequence.
` a
Example : Give first 3 terms of the sequence an = n 2 + n + 1 . Solution : Putting n = 1,2,3 in an = n 2 + n + 1, we get 2 2 2 a1 =1 + 1 + 1 = 3, a2 = 2 + 2 + 1 = 7, a3 = 3 + 3 + 1 = 13 The sequence would be written 3, 7, 13,………
n fffff ffff ffff Example : Give first 3 terms of the sequence an = fffff. 2 n +1 n fffff fffff ffff ffff , we get Solution : Putting n = 1,2,3 in an = 2 n +1 1 2 3 ffff 1f ffff f ffff fffff 2f ffff f ffff fffff 3f ffff ff ffff ff a1 = fffff f, a2 = fffff f, a3 = fffff ff = = = 2 2 2 1 +1 3 2 +1 5 3 + 1 10
1f 2f 3f f f ff ff The sequence would be written as f, f, ff, AAAAAAA 3 5 10
SERIES
If a1 ,a2 ,a3 ,a4 , ……AA ,an , ……… is a sequence, then the expression of their summation a1 + a2 + a3 + a4 + ……A + an + …… is a series. If we take m terms of a finite sequence, the series can be written with summation notation Σ (Sigma notation ).
m
X ak
k=1
= a1 + a2 + a3 + a4 + ……A + am = The sum of all , the value of k being all the natural numbers from 1 to k, k is called the summation variable A
4 1 fffff ffff ffff Example 3 : Write in expanded form : X fffff 3 k=1 k + 1 1 fffff fffff ffff ffff with integers from 1 to 4 and add them Solution : Replace k in 3 k +1 4 1 1 1 1 1f 1f ff ff 1 fffff ffff ffff ffff fffff fffff fffff ffff ffff ffff ffff ffff ffff f f 1f 1f f f ff ff X fffff = fffff fffff fffff fffff = + 3 + 3ffff+ ffff + + ff ff + 3 3 3 2 9 28 65 1 +1 2 +1 3 +1 4 +1 k=1 k + 1
SERIES IDENTITIES:
X ak + X bk = X ak + bk
k=1 n k=1 k=1 n n n
b
c
X ak @ X bk = X ak @ bk X c = cn
k=1 n k=1 k=1
n
n
n
b
c
X cak = c X ak
n
nfffffff n@1 fffffff ffffff X k = fffffff 2 k=1
2 nffffffff n+1 3 ffffffff fffffff X k = ffffffff 4 k=1 n
k=1 n
k=1
`
a
nfff+fff2nffff n f 1 ff + 1 fffffffffffff fffffffffffff X k = ffffffffffffff 6 k=1
2
k=1 n
`
a`
a
`
a2
Xk =
4 k=1
n
nfffffffffffffffffffffffff n + 1 2n + 1 3n 2 + 3n @ 1 fffffffffffffffffffffffff fffffffffffffffffffffffff ffffffffffffffffffffffff
` a`
30
ab
c
ARITHMATIC SEQUENCE & SERIES
A sequence where we start with a number a and repeatedly add a fixed constant d to get the subsequent numbers, is called an Arithmatic Sequence. If we start with 2 and add a fixed constant 3 to it repeatedly, we get the sequence 2, 5, 8, 11, 14, 17, AAAAAAAAAAAA This is an arithmatic sequence. An Arithmatic Sequence is always of the form
a, a + 2d, a + 3d, a + 4d, a + 5d, AAAAAAAAAAA
where we start with the number a and keep on adding d to get the subsequent terms. The number a is the first term (can be written as a1 too) and d is the common difference. To get d in a given arithmetic sequence, subtract any term from its next term. d = a2 @ a1 = a3 @ a2 = a4 @ a3 = ……AA = an @ an @ 1 = ……A
term of an arithmetic sequence is given by The nth ` a an = a + n @ 1 d where an = nth term, a = a1 = first term, d = common difference A
If S n is the sum of n terms of a finite arithmetic sequence, then B ` a nf ` a C nf f f f f S n = f a + a n = f 2a + n @ 1 d 2 2 Example : Write the first 5 terms of the arithmetic sequence 11,8,….. and find its 100th term. Solution : Here a1 = a = 11 and a2 = 8 A So d = a2 @ a1 = 8 @ 11 = @ 3 A Thus each term can be found by adding -3 to the previous term.
Hence the first 6 terms are 11, 8, 5, 2, -1, -4. For the 100th term n = 100, and we have already found a = 11, d = @ 3 ` a Using an = a + n @ 1 d ` a` a a100 = 11 + 100 @ 1 @ 3 ` a = 11 + 99 @ 3 = 11 @ 297 = @ 286 Example : Find the sum of first 12 terms of the arithmetic sequence given by an = 2n + 5 A Solution : Putting n=1 and 2 we get, a = a1 = 2 B 1 + 5 = 7, a2 = 2 B 2 + 5 = 9 A So d = a2 @ a1 = 9 @ 7 = 2 A B ` a C nf f f f Using the summation formula S n = 2a + n @ 1 d 2 B ` a C 12f ff ff f f we get S 12 = 2 B7 + 12 @ 1 2 = 216 A 2
GEOMETRIC SEQUENCE & SERIES :
A sequence where we multiply each term with a non-zero constant to get the next term, is a Geometric Sequence. If we start with 2 and multiply it repeatedly by a non-zero fixed number 3, we get the 2 3 sequence 2, 2 B3, 2 B3 , 2 B3 , AAAAAAAAAAAA or 2, 6, 18, 54, AAAAAAA This is a geometric sequence.
A Geometric Sequence is always of the form
a, ar 2 , ar 3 , ar 4 , ar 5 AAAAAAAAAAAAAA
where we start with the number a and keep on multiplying with r to get the subsequent terms. The number a is the first term (can be written as a1 too) and r is the common ratio. r is the ratio of any term to its previous term, aff aff aff anff f 3 f f 4 f ffff fff f 2 = f= f= = r = ff ff ff ……AA = ffff ……A a1 a2 a3 an @ 1 Examples of Geometric sequence : 1f 1f ff ff f f 1f 1f f f ff ff , , ff ffAAAAAAAAA , , 2 6 18 54 1, @ 10, 20, @ 40, 80, AAAAAAAAAA 2, 0.2, 0.02, 0.002, AAAAAAAAAA
1f 1f f f a = f, r = f 2 3 a = 1, r = @ 5 a = 2, r = 0.1
The nth term of a Geometric Sequence is given by an = a r n @ 1 where an = nth term, a = a1 = first term, r = common ratio A A Geometric Series is the sum of the terms as indicated above. The sum of n terms of it 1fffff @rn fffff fffff is given by S n = a fffffwith r ≠ 1 1@r Example : Write the 6th term and sum of first 6 terms of the geometric sequence 9f f 4, 6, f, AAAAAAAAA 2 Solution : aff 6f 3f ff f f f f f f f Here in this geometric sequence, a = 4, and a2 = 6, the common ratio r = 2 = = . So, a1 4 2
using an = a r
n@1
,
6th term = a6 = 4 B
f g6 @ 1 3f f f
2
243f fff ff ff = fff A 8
1fffff @ r nf ffff fff , Using S n = a fffff 1@r
d e6 3f f f 1@ f fffffff 665f fffffff fff fff2 ff ffffff fff f ff Sum of first 6 terms = S 6 = 4 B = ff 3f f f f 32 1@ 2
Infinite Geometric Series:
An infinite geometric series is of the form a + ar + ar 2 + ar 3 + ar 4 + ………………AA 1fffff @ rn ffff ffff The sum of the series is given by S n = a1 fffff with r ≠ 1 1@r a ffff ffff fff If |r| < 1, then rn gets close to 0 as n gets large. Then, S n gets very close to ffff. 1@r So the sum of the infinite series becomes a ffff ffff ffff S n = ffff when |r| < 1 and n Q 1 1@r Example : Find the sum of the infinite series f gn 8f 16f 32f 2f f ff ff f ff ff f f f f f f 4+ + + + ………+ 4 A + ……… 3 9 27 3
2f f Solution : Here first term = a = 4 and common ratio r = a2 D a1 = f. 3 Using the sum formula of the infinite series a 4 ffff ffff ffff ffff fff ffff S n = ffff= ffff= 12 f f f 1 @ r 1 @ 2f
Therefore, 4 +
3 8f 16f 32f f ff ff f ff ff f f f f
3
+
9
+
27
+ ………1 = 12
MATHEMATICAL INDUCTION
Mathematical induction is a special method of proof which can be used to prove many formulas (like the summation formulas of Arithmatic series, Geometric series etc and many others). First, the proposed formula is declared by a statement P which says something about all natural numbers n. ` a nfff+fff n f1 fffffff fffffff ffffff . Like P(n) : 1 + 2 + 3 + 4 + ……….. + n = 2 1fffff @ rn ffff ffff Or P(n) : a + ar + ar 2 + ar 3 + ………+ ar n @ 1 = a fffff 1@r How do we prove that these will be true for any natural number n? Step 1 : Show that the statement is true when n=1 i.e. P(1) is true. Step 2 : Assume that the statement is true for any natural number k, or P(k) is true. Step 3 : Prove that it is true for the next natural number k+1. That means P(k+1) is true whenever P(k) is true. Then by Principle of Mathematical Induction it is true for all natural numbers n. Explanation : We have proven that P(k+1) is true if P(k) is true and shown that P(1) is true Using the above two we can conclude that as P(1) is true therefore P(1+1) or P(2) is true, as P(2) is true therefore P(2+1) or P(3) is true, as P(3) is true therefore P(3+1) or P(4) is true and so on… So, P(n) is true for all natural number n. Example : Prove by mathematical induction
`
]
a
nfff+fff n f1 fffffff fffffff ffffff 1 + 2 + 3 + 4 + …………+ n = 2
` a
Solution :
nfff+fff n f1 fffffff ffffff Let P(n) denote the statement 1 + 2 + 3 + 4 + …………+ n = fffffff 2 ` a 1ffffff 1fffff +fff 1f fff ff Step 1 : Then P(1) denotes 1 = ff fff which is true. 2 Step 2 : we assume P(k) is true i.e. ` a kfff+fff k f1 fffffff fffffff ffffff 1 + 2 + 3 + 4 + …………+ k = 2 Step 3 : Using step 2 we want to prove P(k+1) is true i.e. ` a` a k ff1 fff+fff ` a fffffffffff fffffffffff fffffffffff f+ f k f 2 1 + 2 + 3 + 4 + …………+ k + k + 1 = 2
To prove this we start with the ` hand side and use step 2 to obtain the right hand side : left a 1 + 2 + 3 + 4 + …………+ k + k + 1 Group the first k terms; @ A ` a Use step 2 or the assumption; = 1 + 2 + 3 + 4 + …………+ k + k + 1 ` a kfff+fff ` k f1 a fffffff fffffff ffffff = + k+1 2 H I
= k+
` ` f g a kf f f J f+ 1 K 1
2
= k+1 =
` a` a k ff1fff+fff fffffffffff fffffffffff fffffffffff f+ f k f 2
f g a kffff +2 ffff ffff ffff
2
2
Thus, P(k+1) is true if P(k) is true. Having proved step 1 & 2, by Principle of Mathematical Induction P(n) is true for all natural numbers n.
Example : Prove by mathematical induction a` a 1f ` 2 2 2 f f 1 + 2 + 3 + ……+ n 2 = n n + 1 2n + 1 6 Solution : a` a 1f ` 2 2 2 f Let P(n) denote the statement 1 + 2 + 3 + ……+ n 2 = fn n + 1 2n + 1 6 a` a 1f ` 2 f f Step 1 : Then P(1) denotes 1 = 1 1 + 1 2 B1 + 1 which is true. 6 Step 2 : we assume P(k) is true i.e. a` a 1f ` 2 2 2 2 f 1 + 2 + 3 + ……+ k = fk k + 1 2k + 1 6 Step 3 : Using step 2 we want to prove P(k+1) is true i.e. CB ` C a2 1f ` aB` a a 2 2 2 2 ` f 1 + 2 + 3 + ……+ k + k + 1 = f k + 1 k + 1 + 1 2 k + 1 + 1 6 To prove this we start with the left hand side and use step 2 to obtain the right hand side : a2 2 2 2 2 ` 1 + 2 + 3 + ……+ k + k + 1 Group the first k terms;
= 1 + 2 + 3 + ……+ k + k + 1
B
2 2 2 2
a` a ` a2 1f ` f = fk k + 1 2k + 1 + k + 1 6 ` a ` aF kffffffff ` 2k + 1 aG ffffffff fffffff fffffff = k+1 + k+1 6
C `
a2
Use step 2 or the assumption;
= k+1
`
` a ` a aF kffffffff fffffff 2k + 1 ffffffff fffffff fffffff k fffff fffffff ffffff f+ 1 6 G
6
+
6
= k+1
`
=
6 ` a` a` a 1f f f = k + 1 k + 2 2k + 3 6 CB ` C ` aB` a a 1f f f = k+1 k+1 +1 2 k+1 +1 6
c ` ab 2 2k fffffffff + 7k + 6 kffffffffffffffff +fffffffffffffff 1fffffffffffff ffffffff ffff ff
2 aF 2kfffkffffffff fffffffffffff fffffffffff6f fffff+ffffff f + f f 6k + fG
6
factorize 2k + 7k + 6 = k + 2 2k + 3
2
`
a`
a
Thus, P(k+1) is true if P(k) is true. Having proved step 1 & 2, by Principle of Mathematical Induction P(n) is true for all natural numbers n.
BINOMIAL THEOREM
An expression in the form a+b is called a binomial. a + 2b, 2x + 3y, a 2 + 2ab etc are binomials. Lets see what form they take when raised to first few integer powers i.e. when the power is 0,1,2,3,4 etc ` a0 a+b =1
`
a+b =a+b
a1 a2 a3
` `
a + b = a 2 + 2ab + b
2 2 3
a + b = a 3 + 3a 2 b + 3ab + b
But what will the expanded form be for higher powers? Like a + b ? It is difficult to expand by multiplying it 15 times. So we take help of the Binomial Theorem :
` a15 `
a+b
an
=
n! r ffffffff ffffffff ffffffff X ffffffffa n @ r b ` a
r=0
n
r! n @ r ! Or
=
n
X
r=0
d e
n n@r r a b r
`
a+b
an
=
d e
n n n n@1 n n@2 2 n n n n@1 a + a b+ a b + …………+ ab + b n 0 1 2 r@1
d e d e d e d e
The Binomial Coefficient When n and r are positive integers and r ≤ n
d e
where k! pronounced k factorial = product of natural numbers from 1 to k as example 5! = 1 B2 B3 B4 B5 = 120, 3! = 1 B2 B3 = 6, 1! = 1 etc ` a Remember 0! = 1 not 0 1. When we expand
`
n! n ffffffff ffffffff ffffffff ` a = ffffffff is called the binomial coefficient; r r! n @ r !
b c
Some characteristics of binomial expansion :
a+b
6+1=7 terms]. ` an 2. In a + b the exponent (power) of a starts at n in the first term and decreases by 1 in each succeeding term down to 0 in the last term. On the other hand the exponent of b starts at 0 in the first term and increases by 1 in each succeeding term up to n in the last term.
Finding any term of an expansion :
n n From the expansion we see that is the binomial coefficient of the 1st term, is 0 1 d e n is the binomial the binomial coefficient of the 2nd term and so on. Therefore, r coefficient of the (r+1)th term, whiche also called the general term. d is n n@ r r # The general term = a r + 1 = a b r
d e d e
an
we get n+1 terms [ expanding
b
2x + 3y
c6
we shall have
To find any term we have to represent it in the form of ar + 1 and use the above formula. d e ` a10 5 The 6th term of a + b will be a6 = a5 + 1 = 10 a10 @ 5 b 5 Example : Calculate Solution :
d e ffffffff fffffffffffff ffffffff fffffffffAffff ffffffff fffffffffffff ff f A 3.4 A fff 7.8 8 = fff8!ffff = 1.2fffff5.6ffff ` a 5! 8 @ 5 ! 5! 3! 5 1.2fffffffffffff ffff fffffffffffffff 6.7fff fffffffffffffff fffff ff A 3.4 A 5.6 A 7.8 fA8 a` a= = `ffffffffffffffff fffff= 56 1.2 A 3.4 A 5 1.2 A 3 1.2 A 3 ` a5
d e
8 5
Example : Use binomial theorem to expand a + b Solution : By binomial theorem
`
2 3 4 5 a + b = 5 a 5 + 5 a 4 b + 5 a 3 b + 5 a 2 b + 5 ab + 5 b 0 1 2 3 4 5 we e calculate that can d d e d e d e d e d e 5 = 1, 5 = 5, 5 = 10, 5 = 10, 5 = 5, 5 =1 0 1 2 3 4 5 So,
a5
d e
d e
d e
d e
d e
d e
`
a + b = a 5 + 5a 4 b + 10a 3 b + 10a 2 b + 5ab + b
a5
2 3 4
5
Example : Write the 8th term of the expansion 3x @ 2y
b
c15
Solution :
b
3x @ 2y
c15
= x + @ 2y
b
D
cE15 e
8th term = a8 = a7 + 1
c 15!fb cb ffff fff ff = ffff x 8 @ 128 y 7 7! 8! b cb c ` a = 6435 x 8 @ 128 y 7
c7 ` a15 @ 7b = 15 x @ 2y 7 d
= @ 823680 x 8 y 7