Question

Expanding $(1+0.2)^{1000}$ by the binomial theorem and doing no further manipulation gives $\begin{aligned}&\binom{1000}{0}(0.2)^0+\binom{1000}{1}(0.2)^1\\&+\binom{1000}{2}(0.2)^2+\cdots\\&+\binom{1000}{1000}(0.2)^{1000}\\&=A_0+A_1+\cdots+A_{1000},\end{aligned}$ 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

Answer: 166

Concepts:binomial theorem inequality optimization

Difficulty rating: 2060

Small Hint:

Compare $A_{k+1}$ directly with $A_k$ instead of estimating the binomial coefficients

Big Hint:

Find the last $k$ for which $\frac{A_{k+1}}{A_k}$ is greater than $1$

Solution:

Consecutive terms satisfy $\frac{A_{k+1}}{A_k}=\frac{1000-k}{k+1}\cdot\frac15.$ This ratio exceeds $1$ exactly when $1000-k>5k+5,$ or $k<\frac{995}{6}.$ Thus the terms increase through $A_{166}$ and decrease afterward. Therefore the largest term is $A_{166}.$

1991 AIME Problem 3

Answer: 166

Difficulty rating: 2060

Small Hint:

Big Hint:

Solution:

Problem 3 in Other Years