Binominal Set and Relations

BINOMIAL SET AND RELATIONS Ex.1 Find the first four terms of the expansion of Sol.  1   ax   bx 2   5  1   ax   bx 2   5 = 5C 0 = (–1) (ax)5 + 5C 1  1 (ax)4   2  bx   +  Sol. 3 4. (–2)4 = 16.8C4 10 8  1   = 8C5 (3x2)3  2x  Sol. Since we require term independent from x  n  – r (   ) = 0 5 As given nC + 4, nC 6 nC 5, nC = 2. 6    = – 189 x 4   T3 = 10C = 10C = 10.9 . 9 2.1 2 2 ( x )8 (3 / x 2 )2 .(– 3 )2.x º = 405 are in AP. nC 5 n! n! n! + = 2 (n  5)! 5! (n  4)! 4! (n  6)! 6!  30 + (n – 5) ( n – 4) Ex.7 If in the expansion of   (n – 7) (n – 14) = 0  n = 7, 14 Sol. by putting x = 1 in the polynomial  3 x2 IJ K 15 the rth term If rth term is independent of x, then by the 15 × 3 – (r–1) ( 3+ 2) = 0  r – 1 = 9  r = 10. Ex.8 If the coefficients of rth and (r +1)th terms in the expansion of (3+7x)29 are equal, then of the expansion of polynomial We get the sum of the coefficients of terms 3 formula Ex.4 Find the sum of the coefficient of the terms (1+ x– 3x2 )2143 FG x H is independent of x, then find r = 2.6 (n – 4) n2 – 21n + 98 = 0 (1+ x– 3x2 )2143 1 1  10 × 2 – r  2  2  = 0  r = 2 i.e. 3rd term. find n Sol. = T5  3  expansion of  x  2  x   of 5th, 6th and 7th terms are in A.P., then  2 = 1120 +........ b3 x 4 4 T8  2 Ex.6 Find the term independent from x in the a2 nC 8C = Ex.3 If in the expansion of (1+ y)n,the coefficient  8  T5 = 8C4(x)4(–2/x)4 a3 a4 2 x + 10 2 – b b x  1 = 56 × (27x6) ×    32 x 5 Sol. (–1) Since (n = 8) is even then there is only one middle term i.e.   1  1  (ax )   2  + 5C3 (ax ) 2   2  +....  bx   bx   2 1   3x   2x   T6 (–1) 2  Ex.5 Find the middle term of the expansion  x   x  3 Ex.2 Find the sixth term in the expansion of Sol. 2143 = (–1) = 1 × –1 = –1. = a5 x 5 – 5 10 2142. [(–1)2]1021. = 2 5C 2 2143  (1+1 – 3) find Sol. We have Tr+1 = 29 Cr329–r (7x) r = (29Cr.329–r.7r) xr BINOMIAL  ar = coefficient of (r +1)th term 29 = Cr. 329–r . 7 r Ex.11 If xm occurs in the expansion of ar = ar–1  = 29  29 Sol. . 329–r.7r r 29 Cr C r1 Cr-1 . 330–r . 7  3 7 . (px)n–3 (1/x)3 = 5/2.  (nC3.pn–3) . xn–6 = 5/2. x0 and n – 6 = 0 nC 3 (p) n–3 n = 6 6C3 = 5/2 = 5/2 Exp. = (1+ x) x4 in the expansion  n (1+ + .....+ xn) nC 1x 2 (2n)! C 2nm  FG 2n  m IJ !FG 4n  m IJ ! H 3 KH 3 K 3 Here Tr+1 = e 2  3 1/ 5 10 j 10 Cr ( 2 ) 10–r (31/5 )r, 1 1 (10–r) and r respectively 2 5  r  10. So both these powers will + nC  sum of required terms = T1 + T11 = x2) n nC x+ nC x2+nC x3+ 1 2 3 (1+ 2n  m 3 be integers together only when r = 0 or 10 of (1+ x + x2 + x3)n 4  r = expansion of and 0 Ex.10 Find the coefficient of 4x x2n–3r of 2 and 3 are  p = 1/2 nC r 2 We observe that in general term Tr+1 powers  p3 = (1/2)3 = (1+ 2nC r where r = 0, 1, 2, ....,10.  p3 = 1/8 Sol. = FG 1 IJ Hx K Ex.12 Find the sum of the rational terms in the Sol. putting n = 6 in it p3 r x2n–r 2n Compairing the coefficient of x and constant term 2nC = (px + 1/x)n 3 Tr+1 = So, coefficient of xm The fourth term in expansion of nC The general term in the expansion of the 2n–3r = m x)n is 5/2 then find the value of n and p T4 = , For the coefficient of xm, we must have Ex.9 If the fourth term in the expansion of (px + 1/ Sol. 2 given expression is r–1 30  r 3 =  r = 21. r 7  2n find the coefficient of xm Now, 29C FG x  1 IJ H xK 2x 4 + ...+ x2n) Coefficient of x4 = n C4+ n C2. n C1+ nC 2 10 C0( 2 )10 + = 32 + 9 = 41 10 C10 (31/5) 10