# Problem Study (Binomial Theorem)

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

### Solution

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

Using binomial expansion,

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

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

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

3p5 = 96

p = 2

r = p6 = 64

s = $\frac{15}{4}$ p4 = $\frac{15}{4}$(2)4 =$\frac{15}{4}$(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   . 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 + $\frac{n\left(n-1\right)}{2}$ an-2 b2 + ... = p + q + r + ...

= an

q  = n an-1

r  $\frac{n\left(n-1\right)}{2}{a}^{n-2}{b}^{2}$

∴

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

$\frac{2n}{n-1}=\frac{32×32}{4×96}$

$\frac{2n}{n-1}=\frac{8}{3}$

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)$\left(x-\frac{2}{x}{\right)}^{6}$.

### = (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

∴

Problems Supported by : Sayar Tun Tun Aung