# The ratio of the coefficient of `x^15` to the term independent of `x` in `( x^2 + ( 2/x) )^15` To solve this problem, we must get familiar with the binomial theorem. The theorem states the following:

`(a+b)^n = ((n),(0)) a^n + ((n),(1))a^(n-1)b + ((n),(2))a^(n-2)b^2 + ... + ((n),(n))b^n`

In our case, we allow `a = x^2` and `b = 2/x`. First, we need to find the conditions under which...

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To solve this problem, we must get familiar with the binomial theorem. The theorem states the following:

`(a+b)^n = ((n),(0)) a^n + ((n),(1))a^(n-1)b + ((n),(2))a^(n-2)b^2 + ... + ((n),(n))b^n`

In our case, we allow `a = x^2` and `b = 2/x`. First, we need to find the conditions under which we have `x^15` and the conditions under which we have a term independent of `x`.

Starting with the former case, we need to find a combination of exponents `m` and `n` so that the following holds true:

`(x^2)^m(2/x)^n = C_15x^15`

Here, `C_15` is simply a constant. Because of exponent rules, we know the following must be true:

`2m - n = 15`

We also know the following based on the binomial theorem:

`m+n = 15`

Solving this system of equations yields the following values for `m` and `n`:

`m = 10`

`n = 5`

Therefore, our term containing `x^15` will have the following:

`(x^2)^10 * (2/x)^5 = 32x^15`

We can't stop here, though! If you look at the binomial theorem, there is another coefficient we need to worry about: that derived from the combination term. If you'll notice from the theorem, the top number in the combination will be the overall number of terms, 15, in our case. The bottom number will be the exponent of the second term, 5, in our case. Our overall term that generates `x^15` will now become:

`((15),(5))*(x^2)^10 * (2/x)^5 = 3003*x^20*32/x^5 = 96096x^15`

Therefore, our coefficient for `x^15` will be 96096.

Now, we must find the term independent of `x`, which will be the constant in this problem. As above, we will use the following term:

`(x^2)^m(2/x)^n`

However, now, the exponents must cancel each other in the following way:

`2m-n = 0`

However, the other condition still holds true, where

`m + n = 15`

Solving this for `m` and `n`, we find:

`m = 5`

`n = 10`

Finding the term including the binomial coefficient the way we did above:

`((15),(10)) * (x^2)^5 * (2/x)^10 = 3003*x^10 * 1024/x^10 = 3075072`

Now, we may find the ratio of the coefficient for `x^15` and the constant term:

`r = 96096/3075072 = 0.03125`

Another way of finding this is to use the exact numbers involved:

`r = ( ((15),(5))*2^5)/( ((15),(10)) * 2^10)`

Now, we can use the fact that `((n),(k)) = ((n),(n-k))` to allow `((15),(5))` and `((15),(10))` to cancel. We can also use the rules of exponents to cancel `2^5` on the top and bottom, leaving us with the following exact result:

`r = 1/2^5 = 1/32 = 0.03125`

I hope this helps!

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