Tiêu đề XI - maths - chapter 8 - BINOMIAL THEOREM-(95-122).pdf ✅

96 NARAYANAGROUP BINOMIAL THEOREM JEE-MAIN-SR-MATHS VOL-I W.E-5: If the coefficients of 3 4 x and x in the expansion of    18 1 1 2     ax bx x in powers of x are both zero, then (a, b) is equal to [MAINS-14] Sol. Coefficient of 3 x  coeff. of 3 x in   18 1 2x   2 coeff.of x in   18 a 1 2x   coeff. of x in   18 b 1 2x  18 3 18 2 18 3 2 1      c .2 a. c .2 b. c .2 0   34 16 17a b ..... 1 3    similarly coeff. of 4 18 4 18 3 18 2 x c .2 a. c .2 b. c .2 0     4 3 2 32a 3b 240.... 2     solve (1) & (2)  a 16  ; 272 b 3  W.E-6:The coefficient of n x in the expansion of 1 1     n x x is Sol. 1 1 1     n     x x x     2 0 1 2 ..... 1 . n n n n n n C C x C x C x      n coefficient n x is     1 1 1 1 .      n n n n C C n n =  1 1    n n W.E-7: The coefficient of 7 x in the expansion of   6 2 3 1   x x x is [AIEEE-2011] Sol. =    6 6 2 1 1   x x coefficient of 7 x =    6 6 6 6      C C C C 1 3 3 2   6 6    C C 5 1    120 300 36 = –144      n n x a x a       2 2 4 4 0 2 4 2 ..... n n n n n n C x C x a C x a           n n x a x a       1 1 3 3 5 5 1 3 5 2 ..... n n n n n n C x a C x a C x a       (i)The number of terms in the expansion of     n n x a x a    is  2 2 n  if n is even,  1 2 n  if n is odd. (ii) The number of terms in the expansion of     n n x a x a    is n/2 if n is even,  1 2 n  if n is odd. W.E-8:The number of terms in the expansion of     20 20 x y x y    2 2 is Sol. The number of terms = 1 2  n (where n=20) = 20 1 11 2   MULTINOMIAL THEOREM  The expansion of   n 1 2 3 r a a a ... a     is called multinomial theorem  The number of terms in   n 1 2 3 r a a a ... a     is   1 1 n r r c n N     · W.E-9: The number of terms in the expansion of (a + b + c +d)n is Sol. Put r = 4 in above formulae (   1 1 n r r c n N     ) No. of terms = ( 1)( 2)( 3) 3! n n n    W.E-10: The number of distinct terms in   3 a b c d e     is Sol. r = 5, n = 3 Number of terms = 1 1    n r Cr =35  The general term in the expansion of  1 2 .... p  n x x x    is 1 2 1 2 1 2 ! .... ! ! ... ! n n np p p n x x x n n n where 1 2 .... n n n n    p

98 NARAYANAGROUP BINOMIAL THEOREM JEE-MAIN-SR-MATHS VOL-I W.E-15:The term independent of x in the expansion of 10 2/3 1/3 1/2 1 1 1                 x x x x x x is Sol.   10 1/2 1/3 1/2 1 1                 x x x [MAINS-2013]     10 10 1/3 1/2 1/3 1/2 1 1         x x x x If  1 th r term is the term independent of x then 10 1/ 3   4 1/ 3 1/ 2    r  The term independent of x is   4 10 10 5 4 4 T C C    1 =210  The term containing the coefficient of mx in the expansion of n p q b ax x        is Tr1 i.e, Tr1= n n r r C a b r  , where np m r p q    (integer)  ‘r’th term from end in the expansion of   n x y  is n r 2    th term from begining. Numerically greatest term: Numerically greatest term in the expansion of (1+ x)n where n is a positive interger. a) If   1 | x | n 1 | x |   = p (an integer), then pth term and (p + 1)th term are the two numerically greatest terms in (1 + x)n Also | t | | t | p  p1 . b) If 1 | x | (n 1)| x |   = p + F where p is an integer and 0 < F < 1, then (p + 1)th term is the numerically greatest term in (1 +x)n . W.E-16: Numerically the greatest term in the expansion of   9 2 3  x when 3 2 x  , is Sol.   9 9 9 3 2 3 2 1 2             x x , r = 6 So, greatest term = 9 6 1 2 .T  = 13 7 3 2  Binomial coefficients: 1 2 , , ,............... C C C C o n n n n n are called the binomial coefficients in the expansion of (x+a)n . They are denoted by co , c1 , c2 , - - - cn respectively. co , c1 , c2 , - - - , cn are the binomial coefficients in the expansion of (1 +x)n . a) 0 1 2 3 ..... 2 n n n n n n n c c c c c       b) 0 1 2 3 ..... ( 1) . 0 n n n n n n n c c c c c        c) 0 2 4 1 3 1 ..... ...... 2 C C C C C n n n n n n         d) 1 1 2 3 2. 3. ... . .2 n n n n n n c c c n c n       e) 1 1 2 2 0 (1 ) 1 ... 2 3 1 ( 1) n c c x cn n c x x x n n x           f) 1 1 2 3 0 2 1 ... 2 3 4 1 1 n n c c c c c n n           g) 1 2 3 0 1 ...... ( 1) 2 3 4 1 1 n n c c c c c n n          h) 2 4 0 2 ... 3 5 1 n c c c n      i) n 1 2 1 ...... 6 C 4 C 2 C n 1 3 5       j) a. C0 + (a+d) . C1 + (a + 2d) . C2 + .......+ (a + nd). Cn = (2a + nd) 2n-1 k) a. C0 - (a+d) . C1 + (a + 2d) . C2 - ......+ (-1)n (a + nd). Cn = 0 l) 2 2 2 2 0 1 2 ............. 2 n Cn C C C C n      m) 2 2 2 2 2 0 1 2 3 ... ( 1) . n n c c c c c       2 / 2 ( 1) , 0, n n n c if n is even if n is odd        n) a 2 2 2 ( ). ( 2 ). 0 1 2 C a d C a d C        2 2 2 .......... . . 2 n n n a nd a nd C C           o) ....... 0 1 1 c c c c r r    + 2n c c c n r n n r          2 ! ! ! n n r n r   