Problem Study (Binomial Theorem)

1.   Given that (p - ${\displaystyle \frac{1}{2}}$ x)⁶ = r - 96x + sx² + ... , find p, r, s.

     Solution

      (p - ${\displaystyle \frac{1}{2}}$ x)⁶ = r - 96x + sx² + ...

      Using binomial expansion,

      ⁶C₀ p⁶ + ⁶C₁ p⁵ (- ${\displaystyle \frac{1}{2}}$ x) + ⁶C₂ p⁴ (- ${\displaystyle \frac{1}{2}}$x)² + ... = r - 96x + sx² + ...

     
        p⁶ + 6 p⁵ (- ${\displaystyle \frac{1}{2}}$ x) + 15 p⁴ (- ${\displaystyle \frac{1}{2}}$x)² + ... = r - 96x + sx² + ...

      p⁶ - 3 p⁵ x + ${\displaystyle \frac{15}{4}}$ p⁴ x² + ... = r - 96x + sx² + ...

      _∴ 3p⁵ = 96

          p = 2

      r = p⁶ = 64

      s = ${\displaystyle \frac{15}{4}}$ p⁴ = ${\displaystyle \frac{15}{4}}$(2)⁴ =${\displaystyle \frac{15}{4}}$(16) = 60.

2.  The 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  ${\displaystyle \frac{q^{2 }}{pr}=\frac{2n}{n-<!--\; -\; -->1}}$ . Given that  
      p = 4, q = 32 and r = 96, evaluate n.

      Solution

      (a + b)ⁿ  = p + q + r + ...

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

      aⁿ + n a^n-1 + ${\displaystyle \frac{n(n-<!--\; -\; -->1)}{2}}$ a^n-2 b² + ... = p + q + r + ...

     ∴ p  = aⁿ 

        q  = n a<sup>n-1

           r  = ${\displaystyle \frac{n(n-<!--\; -\; -->1)}{2}{a}^{n-<!--\; -\; -->2}{b}^2}$ 

     ∴ ${\displaystyle \frac{q^2}{pr}=\frac{{(n{a}^{n-<!--\; -\; -->1})}^2}{a^n\frac{n(n-<!--\; -\; -->1)}{2}{a}^{n-<!--\; -\; -->2}{b}^2} }$

${\displaystyle \; =\frac{2n^2{a}^{2n-<!--\; -\; -->2}{b}^2}{n(n-<!--\; -\; -->1)a^{2n-<!--\; -\; -->2}{b}^2} }$

${\displaystyle \; =\frac{2n}{n-<!--\; -\; -->\mathrm{1 }}}$</sup>

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

          ${\displaystyle \frac{2n}{n-<!--\; -\; -->1}=\frac{32\times <!--\; \times \; -->32}{4\times <!--\; \times \; -->96}}$

          ${\displaystyle \frac{2n}{n-<!--\; -\; -->1}=\frac{8}{3}}$

          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)${\displaystyle (x-<!--\; -\; -->\frac{2}{x}{)}^6}$.

        Solution 

          (x² - 5x + 12)${\displaystyle (x-<!--\; -\; -->\frac{2}{x}{)}^6}$ 

       = (x² - 5x + 12) (${\displaystyle x^6-<!--\; -\; -->6x^5(\frac{2}{x})+15x^4(\frac{4}{x^2})+...}$) 

       = (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 b⁸a⁵ -  ¹²C₆ b⁷a⁶

     By the problem,

     ¹²C₅ b⁸a⁵ - ¹²C₆ b⁷a⁶ = 0 

    ∴ ¹²C₅ b⁸a⁵ = ¹²C₆ b⁷a<sup>6

     ∴ 12C₅ b = 12C₆ a</sup>  
      
    ∴ ${\displaystyle \frac{a}{b}=\frac{{}^{12}{C}_5}{{}^{12}{C}_6} }$
${\displaystyle \; =\frac{{}^{12}{C}_5}{{}^{12}{C}_5\frac{7}{6}} }$
${\displaystyle \; =\frac{6}{7}}$
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