Question 926468
<pre>
If you count the terms starting with 0 instead of 1, 
there are n terms, and the kth term is {{{C(n,k)x^r*y^(n-r)}}}

In the expansion of (x+y)n, the second term is 1620.

That's the first term if you start counting with 0,  So

{{{C(n,1)x^n*y^(n-1)=1620}}}

{{{n*x^n*y=1620}}}

the third term is 4320

That's the second term if you start counting with 0,  So

{{{C(n,2)x^(n-2)*y^2=4320}}}

{{{expr(n(n-1)/2)x^(n-2)*y^2=4320}}} 

{{{n(n-1)x^(n-2)*y^2=8640}}} 

and the fourth term is 5760,

That's the third term if you start counting with 0,  So

{{{C(n,3)x^(n-3)*y^3=5760}}}

{{{expr(n(n-1)(n-2)/6)x^(n-3)*y^3=5760}}}

{{{n(n-1)(n-2)x^(n-3)*y^3=34560}}}

{{{system(n*x^(n-1)*y=1620,n(n-1)x^(n-2)y^2=8640,n(n-1)(n-2)x^(n-3)y^3=34560)}}}

Divide equal by equals. That is,

Divide the sides of the 2nd equation by the sides of the 1st equation

{{{(n(n-1)x^(n-2)y^2)/(n*x^(n-1)*y) =8640/1620}}}

Simplify by canceling and subtracting exponents:

{{{((n-1)y)/x=16/3}}}
{{{3(n-1)y=16x}}}
{{{3(n-1)/16=x/y}}}

Divide the sides of the 3rd equation by the sides of the 2nd equation

{{{(n(n-1)(n-2)x^(n-3)y^3)/(n(n-1)*x^(n-2)*y^2) =34560/8640}}}

Simplify by canceling and subtracting exponents:

{{{((n-2)y)/x=4}}}
{{{(n-2)y=4x}}}
{{{(n-2)/4=x/y}}}

Setting the expressions for {{{x/y}}} equal

{{{3(n-1)/16=(n-2)/4}}}

multiply through by 16

{{{3(n-1)=4(n-2)}}}

{{{3n-3=4n-8}}}

{{{5=n}}}

Taking the first two equations of the system:

{{{system(n*x^(n-1)*y=1620,n(n-1)x^(n-2)y^2=8640)}}}
{{{system(5*x^(5-1)*y=1620,5(5-1)x^(5-2)y^2=8640)}}}
{{{system(5x^4*y=1620,5(4)x^3y^2=8640)}}}
{{{system(5x^4*y=1620,20x^3y^2=8640)}}}
{{{system(x^4*y=324,x^3y^2=432)}}}

Solve the 1st equation for y

{{{y=324/x^4}}}
Substitute in the 2nd
{{{x^3(324/x^4)^2=432}}}
{{{x^3(104976/x^8)=432}}}
{{{104976/x^5=432}}}
{{{104976/x^5=432}}}
{{{104976=432x^5}}}
{{{243=x^5}}}
{{{root(5,243)=x}}}
{{{3=x}}}
Substitute in
{{{y=324/x^4}}}
{{{y=324/3^4}}}
{{{y=324/81}}}
{{{y=4}}}

So x=3, y=4, n=5 

Edwin</pre>