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Mathematics
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Session
Binomial Theorem Session 1
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Session Objectives
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Session Objective
1. Binomial theorem for positive
integral index
2. Binomial coefficients Pascalstriangle
3. Special cases
(i) General term
(ii) Middle term
(iii) Greatest coefficient
(iv) Coefficient of xp
(v) Term dependent of x
(vi) Greatest term
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Binomial Theorem for positiveintegral index
For positive integer n
n n n 0 n n1 1 n n2 2 n 1 n1 n 0 n
0 1 2 n1 na b c a b c a b c a b ... c a b c a b
n n nr r r
r 0
c a b
where
n n
r nr
n! n!c c for 0 r n
r! n r ! n r !r!
are called binomial coefficients. n
r
n n 1 ... n r 1C ,
1.2.3...r
numerator contains r factors
Any expression containing two termsonly is called binomialexpression eg.a+b, 1 + ab etc
Binomial theorem
10 10 107 107 3
10! 10.9.8C 120 C C
7! 3! 3.2.1
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Pascals Triangle
0C0
1C0 C1
1
C2
2
C33
C4
4
C5
5
C2
1
C
3
1
C4
1
C5
1C
5
2C
5
3C
5
4
C4
3
C
3
2
C4
2
2C0
3C0
4C05C0
0
a b 1
1
a b 1a 1b
2 2 2a b 1a 2ab 1b
3 3 2 2 3a b 1a 3a b 3ab 1b
4 4 3 2 2 3 4
a b 1a 4a b 6a b 4ab 1b
5 5 4 3 2 2 3 4 5a b 1a 5a b 10a b 10a b 5ab 1b
n n n 1
r 1 r r c c c
3
4
5
6
10
1
1
1
1
1
1
1
2
3
4
10 5
1 1
1 1
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Observations from binomialtheorem
1. (a+b)n has n+1 terms as 0 r n
2. Sum of indeces of a and b of each termin above expansion is n
3. Coefficients of terms equidistant frombeginning and end is same as ncr =
ncn-r
n
n n 0 n n1 1 n n2 2 n 1 n1 n 0 n0 1 2 n1 na b c a b c a b c a b ... c a b c a b
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Special cases of binomialtheorem
n nn n n n1 n n2 2 n n
0 1 2 nx y c x c x y c x y ... 1 c y
n
r n nr r r
r 0
1 c x y
nn n n n 2 n n n r
0 1 2 n r r 0
1 x c c x c x ... c x c x
in ascending powers of x
n n n n n1 n
0 1 n1 x c x c x ... c
n
n nr
rr 0
c x
n
x 1
in descending powers of x
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Question
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Illustrative Example
Expand (x + y)4+(x - y)4 and hence
find the value of 4 42 1 2 1 Solution :
4 4 4 0 4 3 1 4 2 2 4 1 3 4 0 4
0 1 2 3 4x y C x y C x y C x y C x y C x y
4 3 2 2 3 4x 4x y 6x y 4xy y
Similarly 4 4 3 2 2 3 4x y x 4x y 6x y 4xy y
4 4 4 2 2 4
x y x y 2 x 6x y y
4 4 4 2 2 4Hence 2 1 2 1 2 2 6 2 1 1
=34
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General term of (a + b)n
n n r r
r 1 rT c a b ,r 0,1,2,....,n
n n 01 0r 0, First Term T c a b
n n 1 12 1r 1, Second Term T c a b
n n r 1 r 1r r 1T c a b ,r 1,2,3,....,n
1 2 3 4 5 n n 1
r 0 1 2 3 4 n 1 n
T T T T T T T
kth term from end is (n-k+2)th term frombeginning
n+1 terms
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Question
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Illustrative Example
Find the 6th term in the
expansion of
and its 4th term from the end.
952x
4x
5
Solution :
9 r r9r 1 r
4x 5T C5 2x
4 5 4 59
6 5 1 5 4 5
4x 5 9! 4 5T T C
5 2x 4!5! 5 2 x
39.8.7.6 2 .5
4.3.2.1 x
5040
x
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Illustrative Example
Find the 6th term in the
expansion of
and its 4th term from the end.
9
52x
4x5
Solution :9 r r
9r 1 r 4x 5T C5 2x
4th term from end = 9-4+2 = 7th termfrom beginning i.e. T7
3 6 3 69
7 6 1 6 3 6 3
4x 5 9! 4 5T T C
5 2x 3!6! 5 2 x
3
3
9.8.7 5
3.2.1 x
3
10500
x
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Middle term
CaseI: n is even, i.e. number ofterms odd only one middle term
thn 2
term2
CaseII: n is odd, i.e. number ofterms even, two middle terms
n nn 2 2
n 2 n n1
2 2 2
T T c a b
thn 1
term2
n 1 n 1n 2 2
n 1 n 1 n 112 2 2
T T c a b
thn 3
term2
n 1 n 1n 2 2
n 3 n 1 n 11
2 2 2
T T c a b
Middle term= ?
2n1
xx
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Greatest Coefficientn
rc , 0 r n
CaseI: n even
n1
2
nterm T is max i.e. for r
2Coefficient of middle
n
n
2
C
CaseII: n odd
n 1 n 32 2
n 1 n 1term T or T is max i.e. for r or
2 2
Coefficient of middle
n n
n 1 n 1
2 2
C or C
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Question
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Illustrative Example
Find the middle term(s) in the
expansion of and
hence find greatest coefficient
in the expansion
73x3x
6
Solution :
Number of terms is 7 + 1 = 8 hence 2 middleterms, (7+1)/2 = 4th and (7+3)/2 = 5th
33 4 1347
4 3 1 3 3
x 7! 3 xT T C 3x
6 4!3! 6
13
133
7.6.5 3x 105x
3.2.1 82
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Illustrative Example
Find the middle term(s) in the expansion
of and hence find greatest
coefficient in the expansion
73x3x
6
Solution :
1515
4
7.6.5 x 35x
3.2.1 482 3
43 3 1537
5 4 1 4 4
x 7! 3 xT T C 3x
6 3!4! 6
Hence Greatest coefficient is7 7
4 37! 7.6.5
C or C or 353!4! 3.2.1
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Coefficient of xp in theexpansion of (f(x) + g(x))n
Algorithm
Step1: Write general term Tr+1
Step2: Simplify i.e. separatepowers of x from coefficient andconstants and equate final powerof x to p
Step3: Find the value of r
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Term independent of x in(f(x) + g(x))n
Algorithm
Step1: Write general term Tr+1
Step2: Simplify i.e. separatepowers of x from coefficient andconstants and equate final powerof x to 0
Step3: Find the value of r
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Question
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Illustrative Example
Find the coefficient of x5 in the expansion
of and term independent of x10
2
313x
2x
Solution :
r10 r
10 2r 1 r 31T C 3x 2x
r10 10 r 20 2r 3r
r1
C 3 x2
For coefficient of x5 , 20 - 5r = 5 r = 3
310 10 3 5
3 1 31
T C 3 x2
Coefficient of x5 = -32805
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Solution Cont.
r10 r
10 2r 1 r 31T C 3x 2x
r10 10 r 20 2r 3r
r1
C 3 x2
For term independent of x i.e. coefficient of x0 ,20 - 5r = 0 r = 4
410 10 4
4 1 41
T C 32
Term independent of x
76545
8
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Greatest term in the expansion
Algorithm
Step1: Find the general term Tr+1
Step2: Solve for r1r+1
r
T
T
Step3: Solve for r1r+1r+2
T
T
Step4: Now find the common valuesof r obtained in step 2 and step3
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Question
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Illustrative Example
Find numerically the greatest term(s) inthe expansion of (1+4x)8, when x = 1/3
Solution :
r8
r 1 rT C 4x
r8 rr 1r 18r r 1
C 4xT
T C 4x
8!4x
8 r !r! 9 r 4x8! r
9 r ! r 1 !
36 4r 3r 1 36r
7
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Solution Cont.
36
r 7
r8rr 1
r 18r 2 r 1
C 4xT
T C 4x
8!
8 r !r! r 1
8! 8 r 4x4x
7 r ! r 1 !
3r 3
32 4r
1
29r
7
29 36r
7 7 r = 5 i.e. 6th term
5 58
6 5 1 54 4
T T C 563 3
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Class Test
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Class Exercise 1
Find the term independent of
x in the expansion of
81 1
3 5
1
x x .2
Solution :
8 r r1 1
8 3 5
r 1 r
1
T C x . x2
8 r r8 r8
3 5r
1C x2
40 5r 3r 40 8r 8 r 8 r 8 815 15
r r
1 1C x C x
2 2
For the term to be independent of x 40 8r 0 r 515
Hence sixth term is independent of x and is given by3
86 5
1 8! 1T C . 7
2 5! 3! 8
6T 7
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Class Exercise 2
Solution :
Find (i) the coefficient of x9 (ii) theterm independent of x, in the expansion
of9
2 1x3x
r9 r
9 2r 1 r
1T C x
3x
r9 18 2r r
r
1C x
3
r9 18 3r
r
1C x
3
i) For Coefficient of x9 , 18-3r = 9 r = 3
3 39 9 9 9
4 r
1 9! 1 28T C x x x
3 3! 6! 3 9
hence coefficientof x9 is -28/9
ii) Term independent of x or coefficientof x0,
18 3r = 0 r = 66 6
97 6
1 9! 1 28T C
3 6! 3! 3 243
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Class Exercise 3
Solution :
n n nr r 1 r 2For 2 r n, C 2 C C
n 1 n 1 n 2 n 2r 1 r 1 r r a) C b) 2 C c) 2 C d) C
n n nr r 1 r 2C 2 C C
n n n nr 1 r r 2 r 1C C C C
Now as n n n 1m m 1 m 1C C C
n n n nr 1 r r 2 r 1C C C C
n 1 n 1 n 2r r 1 r C C C
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Class Exercise 4
Solution :
If the sum of the coefficients in theexpansion of (x+y)n is 4096, thenprove that the greatest coefficient inthe expansion is 924. What will be itsmiddle term?
Sum of the coefficients is n 122 4096 2
n 12 i.e. odd number of terms
greatest coefficient will be of the middle term
126
12! 12.11.10.9.8.7C 924
6! 6! 6.5.4.3.2.1
Middle term =12 6 6 6 6
6C x y 924 x y
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Class Exercise 5
Solution :
If
then prove that
n
2 2 2n0 1 2 2n1 x x a a x a x ... a x
n
0 2 4 2n
3 1a a a ... a
2
n2 2 2n0 1 2 2n1 x x a a x a x ... a x ...(i)Replace x by x in above expansion we get
n
2 2 2n0 1 2 2n1 x x a a x a x ... a x ...(ii)
Adding (i) and (ii) we get n n2 21 x x 1 x x 2 4 2n0 2 4 2n2 a a x a x ... a x
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Solution Cont.
n n
2 2
1 x x 1 x x 2 4 2n0 2 4 2n2 a a x a x ... a x
Put x = 1 in above, we get
n 0 2 4 2n1 3 2 a a a ... a
n
0 2 4 2n
3 1a a a ... a
2
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Class Exercise 6
Solution :
Let n be a positive integer. If thecoefficients of 2nd, 3rd, 4th terms in
the expansion of (x+y)n are in AP,then find the value of n.
n n 2 n 3
2 1 3 2 4 3
T C x, T C x , T C x n n n
1 2 3
C , C , C are in AP
n n n2 1 32 C C C
2 n! n! n!
2! n 2 ! 1! n 1 ! 3! n 3 !
1 1 1
n 2 ! n 1 ! 3! n 3 !
6 n 1 n 21
n 2 ! 3! n 1 !
26 n 1 6 n 3n 2 2x 9n 14 0 n 7 n 2 0
n 2 or n 7 n 7
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Class Exercise 7
Solution :
Show that
Hence show that the integral part of
is 197.
6 6
2 1 2 1 198.
6
2 1
6 6
LHS 2 1 2 1
6 5 4 3 2
6 6 6 61 2 3 42 C 2 C 2 C 2 C 2
6 5
6 6 6 65 6 0 1C 2 C C 2 C 2
4 3 2 1
6 6 6 6 62 3 4 5 6C 2 C 2 C 2 C 2 C
6 4 2
6 6 62 4 62 2 C 2 C 2 C
6 62 1 2 1
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Solution Cont.
6 4 2
6 6 62 4 62 2 C 2 C 2 C
3 26! 6!2 2 .2 .2 1
2! 4! 4! 2!
= 2 (8 + 15.4 + 15.2 + 1) = 198 = RHS
Let 6
2 1 I f where
I = Integral part of 6
2 1 and f = fraction part of
62 1 0 f 1 i.e.
6
0 2 1 1 0 2 1 1
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Solution Cont.
f f 198 I Integer
Now as '0 f 1 and 0 f 1
let
62 1 f
6 6
2 1 2 1 I f f 198
'0 f 1 and 0 f 1
f f is an integer lying between 0 and 2
f f 1 I 198 f f 198 1 197
Integer part of is 197. 6
2 1
0 f f ' 2
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Class Exercise 8
Solution :
Find the value of greatest term
in the expansion of20
13 13
Consider
201
1 3
Let Tr+1 be the greatest term r 1 r r 1 r 2T T and T T
r 1 rT T
3 r 120 20
r r 1
1 1
C C3 3
20! 1 20!
.r! 20 r ! r 1 ! 21 r !3
20! 1 20!
.r! 20 r ! r 1 ! 21 r !3
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Solution Cont.
1 1 1.
r 21 r 3
21 r 3 r
21 3 121 21 3 21r
2 23 1
r 1 r 2T T r r 1
20 20r r 1
1 1C C3 3
20! 20! 1
.r! 20 r ! r 1 ! 19 r ! 3
1 1 1.
20 r r 1 3
3 r 3 20 r 20 3 3 120 3 21 3 23
r3 1 23 1
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Solution Cont.
21 3 23 21 3 21
r2 2
6.686 r 7.686
r = 7 is the only integer value lying in this interval
720
8 7
1T 3 C
3
is the greatest term.
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Class Exercise 9
If O be the sum of odd termsand E that of even terms in theexpansion of (x + b)n prove that
n
2 2 2 2O E x b
2n 2n
4OE x b x b
2n 2n2 22 O E x b x b
i)
ii)
iii)
n
x b n n 0 n n 1 1 n 1 n 1 n 0 n
0 1 n 1 nC x b C x b ... C x b C x b
Solution :
n n 0 n n2 20 2O C x b C x b ...
n n1 1 n n3 31 3E C x b C x b ...
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Solution Cont.
n n 0 n n2 20 2O C x b C x b ...
n n1 1 n n3 31 3E C x b C x b ...
n
O E x b O - E = (x-b)n
n n2 2O E O E O E x b x b n2 2x b
4 OE = 2 2 2n 2n
O E O E x b x b
2 22 22 O E O E O E 2n 2nx b x b
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Class Exercise 10
Solution :
In the expansion of (1+x)n the binomialcoefficients of three consecutive termsare respectively 220, 495 and 792, findthe value of n.
Let the terms ber r 1 r 2
T , T , T
n r 1 nr r 1 r 1T C x C 220
n r nr 1 r r T C x C 495
n r 1 nr 2 r 1 r 1T C x C 792
nr r 1
nr 1 r
r! n r !T C n! 220.
T r 1 ! n r 1 ! n! 495C
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Solution Cont.
n
r r 1nr 1 r
r! n r !T C n! 220.
T r 1 ! n r 1 ! n! 495C
r 220 4
n r 1 495 9
9r 4n 4r 4
4n 4r ...(i)
13
Similarly r 1
r 2
T r 1 495 5
T n r 792 8
5 n 88r 8 5n 5r r
13 ...(ii)
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Solution Cont.
4n 4r ...(i)13
5n 8
r ... ii13
From (i) and (ii)
4n 4 5n 8
13 13
4n 4 5n 8
n = 12
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Thank you