Which power of $x$ has the greatest coefficient?

Which power of $x$ has the greatest coefficient?

Question

The power of $x$ which has the greatest coefficient in the expansion of $(1 + \frac{1}{2} x)^{10}$ is

1. $x^2$,

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1. $x^3$,

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1. $x^5$,

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1. $x^{10}$.

Solution

The coefficient of $x^k$, where $1 \leq k \leq 10$ is a whole number, is $c_k ={^{10}C_k}\left(\frac{1}{2}\right)^k = \binom{10}{k}\left(\frac{1}{2}\right)^k=\frac{10!}{k!(10 - k)!} \left(\frac{1}{2}\right)^k.$

1. For $k = 2$, this is $11.25$,

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1. for $k = 3$, this is $15$,

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1. for $k = 5$, this is $7.875$,

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1. for $k = 10$, this is $9.76... \times 10^{-4}$, and so the answer is (b).

Alternatively, look at $\dfrac{c_{k+1}}{c_k}$. We have

$$\frac{c_{k+1}}{c_k} = \frac{10!}{(k+1)!(9-k)!} \times \frac{k!(10-k)!}{10!} \times \frac{2^k}{2^{k+1}} = \frac{10-k}{2(k+1)}.$$

So the $c_k$ are growing when $\dfrac{c_{k+1}}{c_k} > 1$, that is, when $10-k > 2(k+1) \iff k < \frac{8}{3}$, or $k \leq 2$.

So if $k = 2$, then $\dfrac{c_{k+1}}{c_k} > 1$, which means that $c_3 > c_2$.

For $k > 2$, we know $\dfrac{c_{k+1}}{c_k} < 1$, and so $c_k$ decreases from its value at $c_3$.

Therefore $x^3$ has the greatest coefficient, and the answer is (b).