(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-1 k-1x5+ ( k-12+ 2pk-1Rk-1) x4 + (4R k-1 + 2 k-1Pk-1) x3 + (Pk-12 4 k-1) x2 + 4Pk-1x + 4 . = [R2 k-1x3 + 2Rk-1 k-1 x2 + ( 2k-1+ 2Pk-1 Rk-1) x+4R k-1 + 2 k-1Pk-1)] x2 +(Pk-12 4 k-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 +4 k-1 =>P2 k-1 +4(P2 k-1 +4 k-2) Pk-12 + 4 (Pk-22 + 4 k-2) = Pk-12 + 4 Pk-22 + 42 Pk-32 +.............+4k-2P1 + 4k-1 1 => 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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