Problem Study (Binomial Theorem)


1.   Given that (p - 1 2 x)6 = r - 96x + sx2 + ... , find p, r, s.

     Solution

      (p - 1 2 x)6 = r - 96x + sx2 + ...

      Using binomial expansion,

      6C0 p6 + 6C1 p5 (- 1 2 x) + 6C2 p4 (- 1 2 x)2 + ... = r - 96x + sx2 + ...
     
        p6 + 6 p5 (- 1 2 x) + 15 p4 (- 1 2 x)2 + ... = r - 96x + sx2 + ...

      p6 - 3 p5 x 15 4 p4 x2 + ... = r - 96x + sx2 + ...

      3p5 = 96

          p = 2

      r = p6 = 64

      s = 15 4 p4 = 15 4 (2)4 = 15 4 (16) = 60.

2The first three terms in the expansion of (a + b)n in  
      ascending powers of b are denoted by p, q and r 
      respectively. Show that  q 2 p r = 2 n n 1 . Given that  
      p = 4, q = 32 and r = 96, evaluate n.

      Solution

      (a + b)n  = p + q + r + ...

        nC0 an + nC1 an-1 b + nC2 an-2 b2 + ... = p + q + r + ...

      an + n an-1 + n ( n 1 ) 2 an-2 b2 + ... = p + q + r + ...

     = an 

        q  = n an-1

           r  n ( n 1 ) 2 a n 2 b 2  

     ∴ q 2 p r = ( n a n 1 ) 2 a n n ( n 1 ) 2 a n 2 b 2  

                 = 2 n 2 a 2 n 2 b 2 n ( n 1 ) a 2 n 2 b 2  

                = 2 n n 1      

         When p = 4, q = 32 and r = 96, 

          2 n n 1 = 32 × 32 4 × 96

          2 n n 1 = 8 3

          8n - 8 = 6n

         2n = 8

         n = 4

  3.     Using binomial theorem, find the coefficient of x2 
          in the expansion of (3 + x - 2x2)5.

          Solution 

             [3 + (x - 2x2)]5

            = 35 + 5 (34) (x - 2x2) + 10 (33) (x - 2x2)2 + ... 

         = 35 + 405(x - 2x2) + 270 (x2 - 4x3 + 4x4) + ... 

        The coefficient of x2 in the expansion of (3 + x - 2x2)5  

            = 405 (-2) + 270 

            = - 540

4.     Find the coefficient of x4 in the expansion of 

        (x2 - 5x + 12) ( x 2 x ) 6 .

        Solution 

          (x2 - 5x + 12) ( x 2 x ) 6  

       = (x2 - 5x + 12) ( x 6 6 x 5 ( 2 x ) + 15 x 4 ( 4 x 2 ) + . . .

       = (x2 - 5x + 12) (x6 - 12x4  + 60x2 + . . . )

      The coefficient of x4 = 1(60) + 12 (-12) = - 84 

5.    In the expansion of (1 + x)(a - bx)12, the coefficient of
       x8 is zero. Find the value of the ratio a : b.

       Solution

         (1 + x)(a - bx)12 
      = (1 + x)(- bx + a)12
 
      = (1 + x) (12C0 (-bx)12 + 12C1  (-bx)11a +12C2 (-bx)10a2 + 12C3 (-bx)10a3

                           + 12C4 (-bx)9a4 + 12C5 (-bx)8a5+ 12C6 (-bx)7a6 + ...) 
  
     ∴ The coefficient of x8 = 12C5 b8a512C6 b7a6

     By the problem,

     12C5 b8a5 - 12C6 b7a6 = 0 

    12C5 b8a5 = 12C6 b7a6

     12C5 b = 12C6 a  
      
    ∴ a b = 12 C 5 12 C 6  
            = 12 C 5 12 C 5 7 6  
            = 6 7
Problems Supported by : Sayar Tun Tun Aung