Question

If `G(X)=(x^2+x^3+....+x^7)^4`, find the coefficient of `(x^14)`?

Answer 1

`80`

Explanation:

Given:

`g(x)=(x^2+x^3+....+x^7)^4`

Let:

`f(x) = g(x)/x^8 = (1+x+x^2+x^3+x^4+x^5)^4`

The coefficient of `x^14` in `g(x)` is the coefficient of `x^6` in f(x)#.

`(1+x+x^2+x^3+x^4+x^5)^2`

`= (1+2x+3x^2+4x^3+5x^4+6x^5+5x^6+4x^7+3x^8+2x^9+x^10)`

Then:

`(1+2x+3x^2+4x^3+5x^4+6x^5+5x^6+4x^7+3x^8+2x^9+x^10)^2`

`=(...+(1*5+2*6+3*5+4*4+5*3+6*2+5*1)x^6+...)`

That is, the coefficient of `x^6` in f(x)# is:

`1*5+2*6+3*5+4*4+5*3+6*2+5*1`

`= 5+12+15+16+15+12+5 = 80`

Answer 2

80

Explanation:

`G(x)=(x^2+x^3+....+x^7)^4 = x^8(1+x+...+x^5)^4`

So, what we are looking for is the coefficient of `x^6` in

`F(x) = (1+x+...+x^5)^4 = ((1-x^6)/(1-x))^4 = (1-4x^6+...)(1-x)^-4`

Thus, the coefficient we are looking for is

`1times "the coefficient of "x^6" in "(1-x)^-4`
`-4times "the coefficient of "x^0" in "(1-x)^-4`

Now, the coefficient of `x^6` in `(1-x)^-4` is

`((-4)(-5)(-6)(-7)(-8)(-9))/(6!)= 84`

while the coefficient of `x^0` in `(1-x)^-4` is 1

So, the final answer is

`1times 84-4 times 1 = 80`