### Instruction: Expanding $(1+0.2)^{1000}$ by the binomial theorem and doing no further manipulation gives
\[{1000 \choose 0}(0.2)^0+{1000 \choose 1}(0.2)^1+{1000 \choose 2}(0.2)^2+\cdots+{1000 \choose 1000}(0.2)^{1000}= A_0 + A_1 + A_2 + \cdots + A_{1000},\]where $A_k = {1000 \choose k}(0.2)^k$ for $k = 0,1,2,\ldots,1000.$ For which $k$ is $A_k$ the largest?### Response: To answer this question, I need to compare the terms $A_k$ for different values of $k$ and see which one is the biggest.
I notice that $A_k$ is a product of two factors: ${1000 \choose k}$ and $(0.2)^k$.
The first factor, ${1000 \choose k}$, is the number of ways to choose $k$ items out of $1000$, and it has a formula involving factorials: ${1000 \choose k} = \frac{1000!}{k!(1000-k)!}$.
The second factor, $(0.2)^k$, is just a power of $0.2$, which is a fraction less than $1$.
I wonder how these two factors behave as $k$ increases from $0$ to $1000$.
I recall that the binomial coefficients ${1000 \choose k}$ have a symmetric shape: they start at $1$ when $k=0$ or $k=1000$, and they reach a maximum in the middle when $k=500$.
I also recall that the powers of $0.2$ have a monotonically decreasing shape: they start at $1$ when $k=0$, and they approach $0$ as $k$ increases.
So, combining these two factors, I expect that $A_k$ will have a shape that is roughly like a bell curve: it will start at $A_0 = 1$, increase to a maximum somewhere in the middle, and then decrease to $A_{1000} = 0$.
To find the exact value of $k$ where $A_k$ is the largest, I need to compare two adjacent terms and see when they switch from increasing to decreasing.
That is, I need to find the smallest value of $k$ such that $A_k > A_{k-1}$ and $A_k > A_{k+1}$.
To do this, I can divide $A_k$ by $A_{k-1}$ and see when it is greater than $1$.
Using the formula for ${1000 \choose k}$, I get $\frac{A_k}{A_{k-1}} = \frac{1000!}{k!(1000-k)!} \cdot \frac{(k-1)!(1001-k)!}{1000!} \cdot \frac{(0.2)^k}{(0.2)^{k-1}} = \frac{1001-k}{k} \cdot 0.2$.
This fraction is greater than $1$ when $1001-k > 5k$, or equivalently, when $k < \frac{1001}{6}$.
The smallest integer value of $k$ that satisfies this inequality is $k = 166$.
Similarly, I can divide $A_k$ by $A_{k+1}$ and see when it is greater than $1$.
I get $\frac{A_k}{A_{k+1}} = \frac{1000!}{k!(1000-k)!} \cdot \frac{(k+1)!(999-k)!}{1000!} \cdot \frac{(0.2)^k}{(0.2)^{k+1}} = \frac{k+1}{1000-k} \cdot 5$.
This fraction is greater than $1$ when $5k + 5 > 1000 - k$, or equivalently, when $k > \frac{995}{6}$.
The smallest integer value of $k$ that satisfies this inequality is $k = 166$.
Therefore, $k = 166$ is the unique value of $k$ where $A_k$ is the largest, and it is the answer.