$(1+0.2)^{1000}$ by the Binomial theorem gives

Expanding${1000 \choose 0}(0.2)^0+{1000 \choose 1}(0.2)^1+{1000 \choose 2}(0.2)^2+\cdots+{1000 \choose 1000}(0.2)^{1000}$ $= A_0 + A_1 + A_2 + \cdots + A_{1000}$

Where $A_k = {1000 \choose k}(0.2)^k$ for $k = 0,1,2,\ldots,1000$ .

For which $k_{}^{}$ is $A_k^{}$ the largest?

I Thank all my followers!

The answer is 166.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

$\frac{A_{k+1}}{A_{k}}>1\\\Rightarrow \frac{\dbinom{1000}{k+1}×0.2}{\dbinom{1000}{k}}>1\\\Rightarrow \frac{1000-k}{k+1}>5\\\Rightarrow k<165.833$

Therefore, $A_{166}>A_{165}$ but $A_{167}\ngtr A_{166}$ . Thus, maximum is $A_{166}$