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I've noticed that the largest coefficient of $(x+y)^n$ is always the middle one by looking at Pascal's triangle.

I would like to know when the coefficient is maximized in $(2x+y)^n$ or $(x+2y)^n$.

I've looked at the first few $n$'s ($2$, $3$, $4$ and $5$) but I don't see any correlation. Can someone please explain to me?

Edit: using binomial theorem I can get that the $k^{\mathrm{th}}$ coefficient is $\binom n k \cdot 2^k$.

Rócherz
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antoine
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1 Answers1

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There's two approaches.

First, we can take the ratio of two adjacent coefficients. For $(2x+y)^n$, the ratio of the $k^{\text{th}}$ coefficient to the $(k+1)^{\text{th}}$ is:

$$\frac{2^k\binom nk}{2^{k+1} \binom{n}{k+1}} = \frac{2^k\,n!\,(k+1)!\,(n-k-1)!}{2^{k+1}\, n!\,k!\,(n-k)!} = \frac{k+1}{2(n-k)}.$$

So we have $$ 2^k \binom nk \le 2^{k+1}\binom n{k+1} \iff k+1 \le 2(n-k) \iff k \le \frac{2n-1}{3}. $$ So the best coefficient will be roughly $k = \frac23n$. More precisely:

  • When $n$ is divisible by $3$, $k = \frac23n$ is best;
  • When $n \equiv 1 \pmod 3$, $k = \frac{2n-1}{3}$ is tied for best with $k = \frac{2n+2}{3}$;
  • When $n \equiv 2 \pmod 3$, $k = \frac{2n+1}{3}$ is best.

The second approach is less precise and a bit handwavy but gives an intuitive idea of why this happens. Divide by $3^n$, so that we have $(\frac23x + \frac13y)^n$. Then the $k^{\text{th}}$ coefficient is $\binom nk (\frac23)^k (\frac13)^{n-k}$. This is the probability that a biased coin, with probability $\frac23$ of coming up heads, comes up heads $k$ times when you toss it $n$ times.

Intuitively, the likeliest outcome should be that the coin comes up heads about $\frac23$ of the time, so $k \approx \frac23 n$ should maximize the coefficient.

This second approach also gives us an estimate of the maximum coefficient. If $k$ is the likeliest number of heads, then the probability of getting $k$ heads should be at least $\frac1{n+1}$; multiplying by $3^n$, we get $2^k \binom nk \ge \frac{3^n}{n+1}$. On the other hand, the probability is at most $1$ and so $2^k \binom nk \le 3^n$.

This is pretty good for a simple argument; the true value is $O(\frac{3^n}{\sqrt n})$.

Misha Lavrov
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