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?
Solution 1
Let $00$), we have
$\log(A_{k})=\log\left[\frac{(N-k+1)!}{k!}x^{k}\right]=\log\left[\prod_{j=1}^{k}\frac{(N-j+1)x}{j}\right]=\sum_{j=1}^{k}\log\left[\frac{(N-j+1)x}{j}\right]\, .$
Now, $\log(A_{k}^{})$ keeps increasing with $k_{}^{}$ as long as the arguments $\frac{(N-j+1)x}{j}>1$ in each of the $\log\big[\;\big]$ terms (recall that $\log y_{}^{} {\frac{1}{5}}^{k+1}\cdot {{1000} \choose {k+1}}$
Dividing by ${\frac{1}{5}}^k$ gives:
${1000\choose k}>{\frac{1}{5}}\cdot{1000\choose {k+1}}$.
We can express these binomial coefficients as factorials.
$\frac{1000!}{(1000-k)!\cdot(k)!}>{\frac{1}{5}}\cdot\frac{1000!}{(1000-k-1)!\cdot{(k+1)!}}$
We note that the $1000!$ can cancel. Also, $(1000-k)!=(1000-k)(1000-k-1)!$. Similarly, $(k+1)!=(k+1)k!$.
Canceling these terms yields,
$\frac{1}{(1000-k)}>{\frac{1}{5}}\cdot\frac{1}{(k+1)}$
Cross multiplying gives:
$5k+5>1000-k \Rightarrow k>165.8$
Therefore, since this identity holds for all values of $k>165.8$, the largest possible value of $k$ is $\boxed{166}$.

Solution 3
We know that $A_k$ will increase as $k$ increases until certain $k=m$, where $A_0  A_{m+1} > A_{m+2} > \dots > A_{1000}.$
Next, to change $A_{k-1}$ to $A_k$, we multiply $A_{k-1}$ by $\frac{1000-k+1}{5k}$. It follows that the numerator must be greater than the
denominator or
\[1000-k+1>5k\]
\[k<166.8\].
We conclude that the answer is $\boxed{166}$.