17 Feb 2008 | 4 min. (685 words)

  study material-mathematics-algebra-binomial theorem    

Binomial Theorem - 9

(2) If n > 3 then prove that

C0 ab - C1(a - 1) + C2 (a - b) (b - 2) - ……………+ (-1) n Cn = 0

Ans Now, (1 + x)n = 1 + nx + x3 +……….+x put x = - 1

0 = 1 - n + + ……….++ (-1) n ……………….(1)

Now replace n by n - 1

So, 0 = 1 - (n - 1) + +……………..+ (- 1) n-1…………..(2)

Multiplying (1) by a and (2) by n and odding we get.

a - na + a + ……………..+ (-1)na + n - n(n -1) +- …………+ (-1)n-1n = 0

=> a - n (a-1)+ -………..+(-1) n(a-n) = 0………………..(3)

Now Replace n by n-1 and a by a-1 in (3) to get

(a - 1) - (n - 1) (a - 2) + (a - 3) + ……..+ (-1) n-1 (a - n) = 0……(4)

Multiplying (3) by b and (4) by n we get .

ab - n(a - 1) b + (a - 2)b +………..+(-1)n (a - n)b

  • n(a - 1) -n (n - 1)(n - 2)+………..+(-1)n-1 (a - n)n = 0

    On the other hand

    (……(((x - 2)2 - 2 )2 -……….-2)2

    = ((……((x - 2)2 - 2 )2 -……..-2)2-2)2

    k - 1tirmes

    = [ ( Rk-1x3 x2 + Rk-1 x + 4) - 2 ]2

    = ( Rk-1x3 x2 + Rk-1 x + 2)2

    = R2k-1 x6 + 2Rk-1k-1x5+ (k-12+ 2pk-1Rk-1) x4

  • (4R k-1 + 2k-1Pk-1) x3

  • (Pk-12 4k-1) x2 + 4Pk-1x + 4 .

    = [R2 k-1x3 + 2Rk-1k-1 x2 + (2k-1+ 2Pk-1 Rk-1) x+4R k-1 + 2k-1Pk-1)] x2

    +(Pk-12 4k-1) x2 + 4Pk-1x + 4 .

    Whence Pk = 4Pk-1 and k= P2k-1+4 k-1

    Since (x - 2)2 x2 4x + 4

    We have P1 = - 4.

    So, P2 = - 42, P3 = - 43………..

    and so, Pk = - 4k

    NOw let us compute k

    = P2 k-1 +4k-1

    =>P2 k-1 +4(P2 k-1 +4k-2)

    Pk-12 + 4 (Pk-22 + 4 k-2)

    = Pk-12 + 4 Pk-22 + 42 Pk-32 +………….+4k-2P1 + 4k-11

    => ab - n (a - 1)(b - 1) (a - 2)(b - 2)…………+ (-1)n

    Or. C0 ab - C1(a - 1) (b - 1) + C2 (a - 2)(b - 2)………

    +(- 1)n Cn (a - n) b - n) = 0

    (3) Determine coefficients in x2 appeating parantheses have been removed and like terms have been callected is the expression

(.........(( x - 2)2 - 2)2 .........-2 )2

Ans Let us first of all determine constant term which is obtained from the expression.

[ (( x - 2)2 - 2)2 .........-2 ]2

k times

For kthis we put x = 0 i.e it is equal to

(.........((- 2)2 - 2)2 .........-2 )2

k times

(.........(( 4 - 2)2 - 2)2 .........-2 )2

(R -1)times

= (......(4 - 2)2-............-2)2

(R - 2) times

=((4 - 2)2 - 2)2 = (4 - 2)2 = 4

Now let us denote ny Pk the coe fficient of x by Pkthe coefficient of x2 and the Rk the coefficient of x3 the sum of terms involving x to the powers higher than 2 then we can write

(.........(( x - 2)2 - 2)2 -.........-2 )2= Rk x3 + kx2 +Pkx + 4

Now by substituting img ..........= 1, P1 = - 4, P2 =-42.....

into this expression we get,

k= 42k-2 + 4.42k-4+ 42 42k-6+........4k-2 42 + 4k-1 1.

= 42k-2 + 4.42k-3+ 42 42k-4+........4k ++4k-1

= 4k-1 [ 1 + 4 + 42 +..........+4k-1]

=4k-1

=

So, Coefficient of x2 in the given expression is



Dumb Question. What is the significance of the subscript in Rk ,k , Pk etc?

Ans One may note that the expression in a very symetric kind of fexpression where things are repeated k times The subscript k denotes that repetition olnly .

One should always remember that symetry isan important thing in matrematics and should be always used aas an important tool

Key words

(1) Binomial theoram

(2) Binomial Coefficient .

(3) General term .

(4) Multinormial theoram.

(5) Binomial series.

(6) Middle term.

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