Final answer:
To find the coefficient of x^3 y^5 in the expansion of (3x+y)^n, we can use the binomial theorem. By plugging in the values and equating the coefficient to 756k, we find that k = 1.
Step-by-step explanation:
To find the coefficient of x^3 y^5 in the expansion of (3x+y)^n, we can make use of the binomial theorem.
The general term in the expansion is given by C(n,r) * (3x)^(n-r) * (y)^r, where C(n,r) represents the number of ways to choose r objects from a set of n objects.
In this case, we want to find the coefficient when r = 5 and n-r = 3, which will give us the desired term.
Plugging in the values, we have: C(n,5) * (3x)^(n-5) * (y)^5 = 756k. As we can see, we have a coefficient of 756. Now, we need to find the value of k. Since the coefficient is 756k, we can equate it to 756 * k = 756k.
Therefore, the value of k is 1.