Question

find the coefficient of x7 in ( ax2 + 1/bx)11 and the coefficient of x-7 in (ax - 1/bx2)11 . if these coefficients are equal then find the relation between a and b

Answer 1

Answer:

ab = 1

Step-by-step explanation:

Hi,

rth term  in the expansion of (a + b)ⁿ is given by nCr(a)^r(b)^(n-r)

To find the coefficient of x⁷ in (ax² + 1/bx)¹¹

rth term is given by 11Cr(ax²)^r(1/bx)^(11-r)

= 11Cr a^r (1/b)^(11-r) x^(2r -11+r)

=11Cr a^r (1/b)^(11-r) x^(3r -11)

To find the coefficient of x⁷, 3r - 11 should be 7

=> 3r - 11 = 7

=> 3r = 18

=> r = 6

Thus the coefficient of x⁷ is 11C6 a⁶b⁻⁵

To find the coefficient of x⁻⁷ in (ax - 1/bx²)¹¹

rth term is given by 11Cr(ax)^r(-1/bx²)^(11-r)

= 11Cr a^r (-1/b)^(11-r) x^(r -22+2r)

=11Cr a^r (-1/b)^(11-r) x^(3r -22)

To find the coefficient of x⁻⁷, 3r - 22 should be -7

=> 3r - 22 = -7

=> 3r = 15

=> r = 5

Thus the coefficient of x⁷ is 11C5(a)⁵(-1/b)⁶ = 11C6 a⁵b⁻⁶

Given that these coefficients are equal

=>11C6 a⁶b⁻⁵ = 11C6 a⁵b⁻⁶

⇒ a = 1/b

⇒ ab = 1

Hope , it helped !