Question

What is the coefficient of xy4 in the expansion of (2x + y)5?

Answer 1

The coefficient of
`xy^4` in the expension of
`(2x+y)^5` is 10.

Explanation:

Given
`(2x+y)^5` , we have to find the coefficient of
`xy^4` in the expension of
`(2x+y)^5`.

`\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _(i=0)^n\binom{n}{i}a^(\left(n-i\right))b^i`

Here, a = 2x and b = y ,

Substitute above, we get,

`=\sum _(i=0)^5\binom{5}{i}\left(2x\right)^(\left(5-i\right))y^i`

Using
`\binom{n}{i}=(n!)/(i!\left(n-i\right)!)`

`\quad \:i=0\quad :\quad (5!)/(0!\left(5-0\right)!)\left(2x\right)^5y^0`

`\quad \:i=1\quad :\quad (5!)/(1!\left(5-1\right)!)\left(2x\right)^4y^1\\\\\quad \:i=2\quad :\quad (5!)/(2!\left(5-2\right)!)\left(2x\right)^3y^2\\\\\ \quad \:i=3\quad :\quad (5!)/(3!\left(5-3\right)!)\left(2x\right)^2y^3\\\\\quad \:i=4\quad :\quad (5!)/(4!\left(5-4\right)!)\left(2x\right)^1y^4\\\\\quad \:i=5\quad :\quad (5!)/(5!\left(5-5\right)!)\left(2x\right)^0y^5\\\\`

Adding all terms, we get,

`(2x+y)^5=\quad (5!)/(0!\left(5-0\right)!)\left(2x\right)^5y^0+\quad (5!)/(1!\left(5-1\right)!)\left(2x\right)^4y^1+\quad (5!)/(2!\left(5-2\right)!)\left(2x\right)^3y^2+\quad (5!)/(3!\left(5-3\right)!)\left(2x\right)^2y^3+\quad (5!)/(4!\left(5-4\right)!)\left(2x\right)^1y^4+\quad (5!)/(5!\left(5-5\right)!)\left(2x\right)^0y^5`

On evaluating , we get,

`(2x+y)^5=32x^5+80x^4y+80x^3y^2+40x^2y^3+10xy^4+y^5`

Thus, the coefficient of
`xy^4` in the expension of
`(2x+y)^5` is 10.

Answer 2

to solve the questions we proceed as follows:
(2x+y)^5
=(2x+y)^2(2x+y)^2(2x+y)
=(4x^2+4xy+y^2)(4x^2+4xy+y^2)(2x+y)
=32x^5+80x^4y+80x^3y^2+40x^2y^3+10xy^4+y^5
the coefficient of xy^4 is 10
the answer is 10