Find the largest coefficient in this expansion of a binomial

Question

What is the largest coefficient in the expansion of $(x+4)^{10}$?

This is not a homework question, it's a question from a mathleague competition that I did not understand. Please keep answers at the high school mathematics level.

Answer 1

Binomial theorem states that the $k^{th}$ term (if we start counting from $0$) of the expansion is $$\binom{10}{k}\cdot 4^k \cdot x^{10-k}$$

The coefficient of any term is therefore going to be $\binom{10}{k}\cdot 4^k$. We will try to maximize this value. I will do this by calculator:

$k=10, 4^{10}= 1048576, \binom{10}{10}=1 \to 1048576$

$k=9, 4^{9}= 262144, \binom{10}{9}=10 \to 2621440$

$k=8, 4^{8}= 65536, \binom{10}{8}=45 \to 2949120$

$k=7, 4^{7}= 16384, \binom{10}{7}=120 \to 1966080$

$k=6, 4^{6}= 4096, \binom{10}{6}=210 \to 860160$

$k=5, 4^{5}= 1024, \binom{10}{5}=252 \to 258048$

After this, both the binomial coefficients and the power of $4$ will only decrease. Therefore, the largest coefficient is at $k=8$, and is 2949120.

Answer 2

By Binomial Expansion,

$(x+y)^n$ = ${n \choose 0}x^n y^0 + {n \choose 1}x^{n-1}y^1 + {n \choose 2}x^{n-2}y^2 + \cdots + {n \choose n-1}x^1 y^{n-1} + {n \choose n}x^0 y^n$

The problem now boils down to finding the max value of $10\choose k$*$4^k$ as $k$ varies from $0$ to $10$.

Answer 3

More generally, let's look at $(x+c)^n$ where $c$ is a constant.

The expansion is $(x+c)^n =\sum_{k=0}^n \binom{n}{k}x^k c^{n-k} $.

To find the largest term, look at the ratio of consecutive terms:

Let $t(n, k) =\binom{n}{k} c^{n-k} $.

$\begin{array}\\ r(n, k) &=\frac{t(n, k+1)}{t(n, k)}\\ &=\frac{\binom{n}{k+1} c^{n-(k+1)}}{\binom{n}{k} c^{n-k}}\\ &=\frac{\frac{n!}{(k+1)!(n-(k+1))!}c^{-(k+1)}}{\frac{n!}{k!(n-k)!}c^{-k}}\\ &=\frac{\frac{1}{(k+1)}c^{-(k+1)}}{\frac{1}{(n-k)}c^{-k}}\\ &=\frac{n-k}{c(k+1)}\\ \end{array} $

For example, $r(n, 0) =\frac{n}{c} $, $r(n, 1) =\frac{n-1}{2c} $, $r(n, n-1) =\frac{1}{nc} $.

We see that initially, $r(n, k) > 1$ and, eventually, $r(n, k) < 1$. This means that the $t(n, k)$ initially increase and then decrease. This change happens when $r(n, k) \sim 1$; i.e., when $n-k =c(k+1) $ or $n-c =k(1+c) $ or $k =\frac{n-c}{1+c} =k_0(n, c) $.

For $k < k_0(n, c) $, the terms increase, and for $k > k_0(n, c) $, the terms decrease.

(Note that, if $n \le c$, the terms never increase, so the first term is the largest.)

Therefore, the maximum term is within one of $k_0(n, c) $.

As to the precise one, and the possibility that $r(n, k_0(n, c)) = 1 $, those I'll leave for others.