Find the term that contains $b^8$ in the expansion of $(a+b^2)^{12}$

Question

I have tried to plug values into the formula but this is really confusing me for some reason.

Answer 1

Apply the binomial theorem:

$${12 \choose 0}a^{12} + {12 \choose 1}a^{11} (b^2) + {12 \choose 2}a^{10}(b^2)^2 + \cdots+\color{#00A000}{{12 \choose 4}a^8(b^2)^4} + \cdots + {12 \choose 12}(b^2)^{12}$$

Take the term highlighted in $\color{#00A000}{\text{green}}$. We have $${12 \choose 4}a^8(b^2)^4=\frac{12!}{(12-4)!}a^8(b^2)^4=\boxed{495a^8b^8}$$

Answer 2

From binomial theorem $b^8$ appears in the term ${12 \choose 4}a^8(b^2)^4=495a^8b^8$.

Answer 3

Think about how you would actually expand the expression. Written out, you have $12$ sets of parentheses. You can get a $b^8$ term by combining four $b^2$ terms with eight $a$'s. Hence, the ultimate term will be $a^8b^8$. Further, there are $\binom{12}{4}$ ways of selecting four $b^2$ terms from the $12$ sets of parentheses.

Therefore, the term and its coefficient will be $\binom{12}{4}a^8b^8 = 495a^8b^8$. No formula memorization required, except for maybe how to expand $\binom{n}{k}$. (Ultimately, it's just the binomial theorem though: see Mario's answer).