Question

Find the coefficient of 

(i) x⁵ in the expansion of (x + 3)⁸

(ii) x⁶ in the expansion of \(\Big(3x^2-\frac{1}{3x}\Big)^9\) 

(iii) x^-15 in the expansion of \(\Big(3x^2-\frac{a}{3x^3}\Big)^{10}\)

(iv) a⁷b⁵ in the expansion of (a – 2b)¹² .

Answer 1

(i) Here, a=x, b=3 and n=8

We have a formula,

To get coefficient of x⁵ we must have, 

(x)^8-r = x⁵ 

• 8 - r = 5 

• r = 3

Therefore, coefficient of x⁵ = (⁸₃)(3)³

=  \(\frac{8\times7\times6}{3\times2\times1}\) .(27)

= 1512 

(ii) Here, a=3x² , b = \(\frac{-1}{3x}\)  and n = 9 

We have a formula,

To get coefficient of x6 we must have,

(x)^18-3r = x⁶ 

• 18 - 3r = 6 

• 3r = 12 

• r = 4

Therefore, coefficient of x⁶ =  (⁹₄)(3)^9-4\(\Big(\frac{-1}{3}\Big)^4\)

\(\frac{9\times8\times7\times6}{4\times3\times2\times1}\) .(3)⁵ \(\Big(\frac{1}{3}\Big)^4\)

= 126 × 3 

= 378

(iii) Here, a = 3x²,b = \(\frac{-a}{3x^3}\) and n = 10

We have a formula,

To get coefficient of x^-15 we must have, 

(x)^20-5r = x - 15 

• 20 - 5r = -15 

• 5r = 35 

• r = 7

Therefore, coefficient of x^-15 = (¹⁰₇)(3)^10-7 \(\Big(\frac{-a}{3}\Big)^7\) 

(iv) Here, a = a, b = - 2b and n = 12 

We have formula,

To get coefficient of a⁷b⁵ we must have,

(a)^12-r (b)^r = a⁷b⁵ 

• r = 5

Therefore, coefficient of a⁷b⁵ = (¹²₅)(-2)⁵

= \(\frac{12\times11\times10\times9\times8}{5\times4\times3\times2\times1}\) .(-32)

= 792. (-32) 

= -25344