Ratio of binomial coefficients If the ratio of the coefficients of $x^6$ and $x^7$ in the expansion of $(2+ax)^{11}$ is $14:25$, find the value of $a$.
What I have so far:
$$\binom{11}{6} \cdot a^6 \cdot 2^5$$
And
$$\binom{11}{7} \cdot a^7 \cdot 2^4$$ 
 A: You calculated the coefficients correctly. If $A$ is the coefficient of $x^6$ and $B$ is the coefficient of $x^7$, then you want to find $A/B$, and set it equal to $14/25$. 
You calculated the following:
$$A = \binom{11}{6} \cdot a^6 \cdot 2^5$$
$$B = \binom{11}{7} \cdot a^7 \cdot 2^4$$
We can write out the $\binom{11}{6}$ and $\binom{11}{7}$ terms to get:
$$A = \binom{11}{6} \cdot a^6 \cdot 2^5 = \binom{11}{5} \cdot a^6 \cdot 2^5$$
$$A = \frac{11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{5 \cdot 4 \cdot 3 \cdot 2} \cdot a^6 \cdot 2^5$$
$$B = \binom{11}{7} \cdot a^7 \cdot 2^4 = \binom{11}{4} \cdot a^7 \cdot 2^4$$
$$ B = \frac{11 \cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2} \cdot a^7 \cdot 2^4$$
Now we can calculate $A/B$:
$$\frac{A}{B} = \frac{\frac{11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{5 \cdot 4 \cdot 3 \cdot 2} \cdot a^6 \cdot 2^5}{\frac{11 \cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2} \cdot a^7 \cdot 2^4}$$
Lots of things conveniently cancel and we end up with
$$\frac{A}{B} = \frac{\frac{7}{5} \cdot a^6 \cdot 2^5}{a^7 \cdot 2^4}$$
$$\frac{A}{B} = \frac{7 \cdot 2}{5 \cdot a}$$
We want this to equal $14/25$.
$$\frac{A}{B} = \frac{14}{25} = \frac{7 \cdot 2}{5 \cdot a}$$
$$\frac{14}{25} = \frac{14}{5a}$$
$$25 = 5a$$
$$\boxed{a=5}$$
