What is the greatest term in $(1+\frac{1}{2}x)^{12}$ when $x=\frac{1}{2}\;?$

Question

The question is find the greatest term in $\bigg(1+\dfrac{1}{2}x\bigg)^{12}$ when $x=\dfrac{1}{2}$.

I know how to expand using binomial expansion but there no background in my book related to this question, the only thing that I know to expand it and find out the total 13 terms and then pick the greatest but term but that is not the way as I think.

Answer 1

Write

$$\left(1+\frac14\right)^{12}=\sum_{k=0}^{12}\binom{12}k4^{-k}$$

But we have

$$\binom{12}{k+1}4^{-(k+1)}\le\binom{12}k4^{-k}\implies\frac{12!}{(k+1)!(12-k-1)!}4^{-k-1}\le\frac{12!}{k!(12-k)!}4^{-k}\iff$$

$$\frac{(12-k)!}{4(12-k-1)!}\le\frac{(k+1)!}{k!}\iff\frac{12-k}4\le k\iff5k\ge12\iff k\ge3$$

Thus the sequence $\;\left\{\binom{12}k4^{-k}\right\}_{k=0}^{12}\;$ is monotonic descending for $\;k\ge 3\;$, so...what can you deduce from here.

Answer 2

I very much like DonAntonio's answer, however for this simple example I think your textbook expected a solution along the following lines:

Each term in the expansion is $$\binom {12}k\Big(\frac{1}{2}x\Big)^k = \binom {12}k\Big(\frac{1}{4^{k}}\Big)$$ when $x=\tfrac 12$.

Now $4^{k}$ is increasing and grows faster than $\binom{12}k$, so we just need to find when $\binom{12}k/4^k$ first decreases: \begin{align} \begin{array}{c|cc|c} k&4^k & \binom{12}k & \binom{12}k/4^k\\ \hline 0 & 1&1 & 1\\ 1 & 4 & 12 & 3\\ 2 & 16 & 66 & 4.1\\ 3 & 64 & 220& 3.4 \end{array} \end{align} Thus we get a decrease when $k=3$, and so the largest term occurs when $k=2$.