1991 AIME Problem 3

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

Compare

A_{k+1}

directly with

A_k

instead of estimating the binomial coefficients

Find the last

k

for which

\frac{A_{k+1}}{A_k}

is greater than

1

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}.

Problem 3 in other years