Problem: Expanding $(1+0.2)^{1000}$ by the binomial theorem and doing no further manipulation gives $ \begin{gathered} \binom{1000}{0}(0.2)^0+\binom{1000}{1}(0.2)^1+\binom{1000}{2}(0.2)^2+\cdots+\binom{1000}{1000}(0.2)^{1000} \\ =A_0+A_1+A_2+\cdots+A_{1000} \end{gathered} $ where $A_k=\binom{1000}{k}(0.2)^k$ for $k=0,1,2, \ldots, 1000$. For which $k$ is $A_k$ the largest? Answer: We need to find the smallest $k$ such that $ \frac{1}{5^k }\cdot\binom{1000}{k}>\frac{1}{5^{k+1}} \cdot\binom{1000}{k+1} $ Dividing by $\frac{1^k}{5}$ gives: $ \binom{1000}{k}>\frac{1}{5} \cdot\binom{1000}{k+1} $ and we find $k > 165.8$ so $k = 166$