when I do this question, I try not using the: $(n-k+1)/k * b/a$ formula, but rather the $T(k+1)/T(k) ≥ 1$ formula. However, when I do it like that, I get the wrong answer – which is probably a simple algebraic mistake.
The question was:
"Find the greatest coefficient in the expansion of $(1/3 +2x)^{18}$"
So can someone please, step by step, find the greatest coefficient using the $T(k+1)/T(k) ≥ 1$ formula?
Thanks
Best Answer
Note that our expression is equal to $\frac{1}{3^{18}}(1+6x)^{18}$.
The coefficient of $x^{k+1}$ is $\binom{18}{k+1}6^{k+1}$. The coefficient of $x^k$ is $\binom{18}{k}6^{k}$. Divide. We get some immediate cancellation of powers of $6$, and of $18!$, and end up with the preliminary simplification $$6\frac{k!(18-k)!}{(k+1)!(17-k)!}$$ and then $$6\frac{18-k}{k+1}.$$ We want to solve the inequality $6\frac{18-k}{k+1}\le 1$. Manipulation gives the equivalent $k\ge \frac{107}{7}\approx 15.3$. So the maximum coefficient is the coefficient of $x^{16}$.