ဒီေန႔ group မွာ ဆရာ ကိုၿဖိဳး ဆရာ ေက်ာ္ေဇျမင့္ တို႔ရဲ့ မွတ္သားဖြယ္ရာ ပုစာၦေတြကို အေျဖႏွင့္တကြ တင္ေပးလိုက္ပါတယ္။ ကၽြန္ေတာ္စဥ္စားမိသလို တြက္ေပးထားတာ ျဖစ္တဲ့အတြက္ ဒီထက္ပိုမို အဆင္ေျပ ေကာင္းမြန္တဲ့ တြက္နည္းမ်ား ရွိမွာ အမွန္ပါ။ ေက်ာင္းသားမ်ားအတြက္၊
ေလ့လာသူအတြက္ တစ္စံုတရာ
အက်ိဳးရွိေစမယ္လို႔ ယံုၾကည္လ်က္....။

 John Napier (inventor of logarithms)

Logarithm ကို အေျခခံက်က်ေလး ရွင္းျပေပးထားလို႔ နားလည္လြယ္မွာပါ။ ေလ့က်င့္ဖို႔ exercises ေတြလည္းပါေတာ့ အသံုး၀င္မယ္လို႔ ေမွ်ာ္လင့္ပါတယ္။


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