Find the coefficient of xy^8 in the expansion of (x-y^2)^5?

2 Answers
May 17, 2017

Coefficient of xy^8 is 5

Explanation:

Binomial expansion of (a-b)^n is given by

C_0^na^n(-b)^0+C_1^na^(n-1)(-b)^1+C_2^na^(n-2)(-b)^2+C_3^na^(n-3)(-b)^3+......+C_r^na^(n-r)(-b)^r+..........+C_n^na^0(-b)^n, where C_r^n=(n!)/((n-r)!r!).

Hence (r+1)^(th) term is C_r^na^(n-r)(-b)^r

As we are given (x-y^2)^5, its (r+1)^(th) term is

C_r^5x^(5-r)(-y^2)^r

As we are seeking coefficient of xy^8, it means r=4

and corresponding term is C_4^5x^1(-y^2)^4

= 5xy^8 and coefficient of xy^8 is 5

May 17, 2017

The coefficient is 5.

Explanation:

The question is asking what is the coefficient associated with the term xy^8 in the expansion of (x - y^2)^5. To make it easy on myself I used Pascal's triangle to help me in the expansion. The triangle has many interesting properties, but for our purposes we need to realise that each row gives the coefficients for the expansion of the general expression (a + b)^n

(a+b)^0 1
(a + b)^1 1 1
(a+ b)^2 1 2 1
(a + b)^3 1 3 3 1
(a + b)^4 1 4 6 4 1
(a + b)^5 1 5 10 10 5 1
... and so on
So in our case if you let a = x and b = y^2
(x - y^2)^5 = x^5 - 5x^4y^2 + 10x^3y^4 - 10x^2y^6 + 5xy^8 - y^10
From this just read off your answer.