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Basic Concepts Of Binomial Expression Part 2

Posted on - 04-01-2017

Math

IIT JEE

Properties of Binomial  Coefficient

For the sake of convenience the coefficients nC0 , nC1 , . . . , nCr , . . . ,nCn are usually denoted by C0 , C1 , . . . , Cr , . . . ,Cn respectively .

  • C0 + C1 + C2 +. . . . . + Cn = 2n.
  • C0 - C1 + C2 -. . . . . Cn = 0.
  • C0 + C2 + C4 +. . . . . = C1 + C3 + C5 +. . . . . = 2n-1.
  • nCr + nCr-1 = n+1Cr
  • r nCr =n n-1Cr-1
  • .
  • If nCx = nCy, then either x = y or x + y = n.

    So, nCr = nCn - r =

    Some Important results

    (i) Differentiating (1 + x)n = C0 + C1x + C2x2 + . . . + Cnxn , on both sides we .

    have, n(1 + x)n-1 = C1 + 2C2x + 3C3x2 + . . . + nCnxn-1 . . . (1).

    x = 1
    &⇒
    n2n-1 = C1 + 2C2 + 3C3 +
    . . . +nCn.

    x = -1
    &⇒
    0 = C1 - 2C2 +
    . . . +(-1)n-1 nCn.

    Differentiating (1) again and again we will have different results.

    (ii) Integrating (1 + x)n , we have,

    + (where C is a constant )

    Put x = 0, we get C = -

    Therefore …(1)

    Put x = 1 in (1) we get

    Put x = -1 in (1) we get,

    (i) Problems Related to series of Binomial coefficients in which each term is a product of an integer and a binomial coefficient i.e. in the form k nCr .

    Example.1

    If (1+x)n = then prove that C1 + 2C2 + 3C3+ . . . + nCn= n2n-1

    Solution

    Method (i): By summation

    rth term of the given series, tr = r nCr
    &⇒
    tr = n n-1Cr-1

    Sum of the series = =

    = = n 2n-1 .

    Method (ii) By calculus

    We have ( 1+ x )n = C0 + C1 + C2 x2 + . . . + Cnxn …(1).

    Differentiating (1) with respect to x

    n(1 +x )n-1 = C1 +2C2x + 3C3 x2 + . . . + n Cnxn-1 …(2).

    Putting x = 1 in (2)

    n 2n-1 = C1 + 2C2 + . . . + nCn .

    (ii) Problems related to series of binomial coefficients in which each term is a binomial coefficient divided by an integer i.e. in the form of .

    Example.2

    If (1+ x)n = , show that C0 +

    Solution

    Method (i): rth term of the given series

    tr =

    Required sum = =

    = = .

    Method (ii): By calculus

    (1+x)n = C0 + C1x + C2x2 + . . + Cnxn . . . . (1).

    Integrating both the sides of (1) with respect to x between the limits 0 to x


    &⇒
    . . . . (2)

    Substituting x = 1 in (2) , we get

    .

    (iii) Problem related to series of binomial coefficients in which each term is a product of two binomial coefficients.

    Example.3

    If ( 1+x)n = then prove that

    C0C1 + C1C2 + . . . . + Cn-1 Cn = .

    Solution

    We have

    (1+x )n = C0 +C1 x +C2x2 + . . + Cnxn …. (1).

    and ( x+1)n = C0xn +C1xn-1 + . . . + Cn ….(2).

    multiplying (1) and (2), we get

    (1+x )2n = (C0 +C1 x +C2x2 + . . + Cnxn ) (C0xn +C1xn-1 + . . . + Cn).

    Equating the coefficient of xn+1, we get

    C0C1 +C1C2 + . . . + Cn-1Cn = 2nCn+1 = .

    Example.4

    If ( 1+x)n = then prove that .

    Solution

    We have

    (1 – x)2n = 2nC0 - 2nC1 x + 2nC2 x2 - . . . . + (-1 )2n. 2nC2nx2n ….(1).

    and also ( x + 1)2n = 2nC0 x2n + 2nC1 x2n-1 +. . . + 2nC2n ….(2).

    multiplying (1) and (2), we get (2nC0 – 2nC1 x + 2nC2 x2 + . . . . + ( -1 )2n. 2nC2nx2n )( 2nC0 x2n + 2nC1 x2n-1 +. . . + 2nC2n) = ( 1 – x2)2n equating the coefficient of x2n, we get .

    (iv) Questions involving alternate binomial coefficients:

    Example.5

    Evaluate C0 - C1 + C2 - C3 +...+ (-1)nCn.

    Solution

    Here alternately +ve and - ve sign occur

    This can be obtained by putting (-1) instead of 1 in place of x in

    (1 + x)n = C0 + C1x +...+ nCnxn, we get C0 - C1 +...+ (-1)nCn = 0.

    Now to obtain the sum C0 + C2 + C4 + ...we add (1 + 1)n and (1 - 1)n.

    Similarly, the cube roots of unity may be used to evaluate

    C0 + C3 + C6 + ... OR C1 + C4 +... OR C2 + C5 +...

    put x = 1, x = w, x = w2 in

    (1 + x)n = C0 + C1x +...+ Cnxn and add to get C0 + C3 + C6 +... the other two may be obtained by suitably multiplying (1 + w)n and .

    (1 + w2)n by w and w2 respectively.

    Example.6

    Prove that C0 + C4 + C8 + L =

    Solution

    Since (1+i)n = C0 + C1i + C2i2 + C3i3 + L + Cnin

    = (C0 - C2 + C4 -L ) + i(C1 - C3 + L)

    But (1 + i)n =

    Hence = C0 - C2 + C4 + L - - - (1)

    Also 2n-1 = C0 + C2 + C4 + L - - - (2)

    Adding (1) and (2) we have finally

    C0 + C4 + C8 + L =

    (v) Questions involving the greatest integer function:

    These questions generally involve working with binomial expansions on surds.

    Example.7

    If I is integral part of (2 + )n and f is fraction part of (2 + )n, then prove that (I + f) (1 –f) = 1. Also prove that I is an odd Integer.

    Solution

    (2 + Ö3)n = I + f where I is an integer and 0 £ f < 1 show that I is odd and that (I + f) (1 - f)=1

    Here note that (2 - Ö3)n (2 +Ö3)n = (4 - 3)n = 1

    \ (2 + Ö3)n (2 -Ö3)n = 1 it is thus required to prove that

    (2 - Ö3)n = 1 – f but, (2 - Ö3)n + (2 +Ö3)n = [2n - C1.2n - 1.Ö3 + C22n - 2..(Ö3)2 - ...]
    + [2n + C1
    .2n - 1.Ö3 + C22n -2..(Ö3)2 - ...] = 2[2n + C2.2n - 2.3+C42n - 4.32 + ...].

    = even integer

    \ Now 0 < (2 - Ö3) < 1

    \ 0 < (2 - Ö3)n < 1

    \ if (2 - Ö3)n = f '

    then I + f + f ' = Even

    Now O £f < 1 and 0 < f ' < 1 . . . (1).

    Also I + f + f ' = Even integer

    \ f + f ' = Even integer . . . (2).

    (1) and (2) imply that f + f ' = 1 (Q 0 < f + f ' < 2)

    \ I is odd and f ' = 1 - f
    &⇒
    (I + f) (1 - f) = 1.

     
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