Question

The coefficient of x^3y^4 in (3x+2y)^7 is

Answer 1

Coefficient of
`x^3y^4` in
`(3x+2y)^7` is 15120

Explanation:

We know that
`(x+y)^(n)`) can be expanded in (n+1) terms by using binomial theorem and each term is given as

`n_C_(r)x^(n-r)y^(r)`

Here value of r is taken from n to 0

we have to determine the coefficient of
`x^3y^4` in
`(3x+2y)^7`

in this problem we have given n=7

We have to determine the coefficient of
`x^3y^4`

it means in the expansion we have to find the the 3rd power of x and therefore

r=n-3

here n=7

therefore, r=7-3=4

Hence the coefficient of
`x^3y^4` can be determine by using formula

`n_C_(r)x^(n-r)y^(r)`

here n=7, r=4

`7_C_(4)x^(7-4)y^(4)`

=
`(7* 6* 5* 4)/(1* 2* 3* 4) (3x)^3(2y)^4`

=
`15120x^3y^4`

Therefore the coefficient of
`x^3y^4` in
`(3x+2y)^7` is 15120

Answer 2

The coefficient is 15120.

Explanation:

Since, by the binomial expansion formula,

`(x+y)^n=\sum_(r=0)^n^nC_r x^(n-r) y^r`

Where,
`^nC_r=(n!)/(r!(n-r)!)`

Thus, we can write,

`(3x+2y)^7 = \sum_(r=0)^n ^7C_r (3x)^(7-r) (2y)^r`

For finding the coefficient of
`x^3y^4`,

r = 4,

So, the term that contains
`x^3y^4` =
`^7C_4 (3x)^3 (2y)^4`

`=35 (27x^3) (16y^4)`

`=15120 x^3 y^4`

Hence, the coefficient of
`x^3y^4` is 15120.