24. Binomial Theorem-1 (1)

download 24. Binomial Theorem-1 (1)

of 48

Transcript of 24. Binomial Theorem-1 (1)

  • 7/28/2019 24. Binomial Theorem-1 (1)

    1/48

    Mathematics

  • 7/28/2019 24. Binomial Theorem-1 (1)

    2/48

    Session

    Binomial Theorem Session 1

  • 7/28/2019 24. Binomial Theorem-1 (1)

    3/48

    Session Objectives

  • 7/28/2019 24. Binomial Theorem-1 (1)

    4/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    5/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    6/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    7/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    8/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    9/48

    Question

  • 7/28/2019 24. Binomial Theorem-1 (1)

    10/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    11/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    12/48

    Question

  • 7/28/2019 24. Binomial Theorem-1 (1)

    13/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    14/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    15/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    16/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    17/48

    Question

  • 7/28/2019 24. Binomial Theorem-1 (1)

    18/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    19/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    20/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    21/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    22/48

    Question

  • 7/28/2019 24. Binomial Theorem-1 (1)

    23/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    24/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    25/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    26/48

    Question

  • 7/28/2019 24. Binomial Theorem-1 (1)

    27/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    28/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    29/48

    Class Test

  • 7/28/2019 24. Binomial Theorem-1 (1)

    30/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    31/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    32/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    33/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    34/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    35/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    36/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    37/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    38/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    39/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    40/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    41/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    42/48

    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.

  • 7/28/2019 24. Binomial Theorem-1 (1)

    43/48

    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 ...

  • 7/28/2019 24. Binomial Theorem-1 (1)

    44/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    45/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    46/48

    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)

  • 7/28/2019 24. Binomial Theorem-1 (1)

    47/48

    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

  • 7/28/2019 24. Binomial Theorem-1 (1)

    48/48

    Thank you