In the expansion of $a{x}^3 {(2+ax)}^{11}$ , the coefficient of the term in $x^5$ is $11880$. Find the value of $a$ .

Question

This is a question from the ib question bank which has the answer however the method isnt explained well enough. Its binomial theorem. Can someone explain the method in detail of how to deal with these sums when theres a value outside?

Answer 1

Hint

The coefficient of $x^2$ in $(2+ax)^{11}$ is $\binom{11}{2} \cdot a^2\cdot 2^9.$

The coefficient of $x^5$ in $ax^3(2+ax)^{11}$ is $\binom{11}{2} \cdot a^3\cdot 2^9.$

Answer 2

$$(2+ax)^{11}=2^{11}+11\times2^{10}ax+\binom{11}{2}2^9(ax)^2+\cdots$$

Therefore $$ax^3(2+ax)^{11}=2^{11}ax^3+22\times2^{10}a^2x^4+\binom{11}{2}2^9a^3x^5+\cdots$$ Thus you're given that $\frac{11\times10\times 2^9}{2}a^3 = 11880$, which gives $a=3/4$.

Answer 3

Hint:

You can just simply "put it aside" until you deal with the binomial expression.

$$ (2+ax)^{11}=\sum_{k=0}^{11}\binom{11}{k}2^{11-k}\cdot (ax)^k $$

from here (since you will later multiply this by $ax^3$) you will need the term which raises $(ax)$ to the $2$nd power.

Hope this helps :)