What is the coefficient of$ x^6y^1$ in the expansion of $(3x^2+y)^4$?

Question

So this is what I have so far. $(3x^2)^4$ + $\binom{4}{1}(3x^2)^3(y)$

Why is the answer not 4? How do I continue?

Answer 1

$\binom{4}{1}(3x^2)^3 y = 4(3^3)(x^6)y = 108x^6y$ thus the coefficient is $108$

Answer 2

Recall that

$$(a+b)^4=\sum_{k=0}^4{4\choose k}a^kb^{4-k}$$ so in your case we take $k=3$ to get the desired coefficient: $3^3{4\choose 3}=27\times 4$

Answer 3

Note that $$(3x^2+y)^4 = \sum_{j=1}^4\binom{4}{j}(3x^2)^{4-j}y^j$$ Now we can pattern match to what you have. We are clearly interested in the case of $j=1$. Thus we consider $$\binom{4}{1}(3x^2)^3y =4\cdot 3^3x^3y \\ = 108x^3y$$