1971 AMC 12 Problem 24

Question

Pascal’s triangle is an array of positive integers, shown below, in which the first row is

1,

the second row is two

1

’s, each row begins and ends with

1,

and each other entry is the sum of the two entries above it.

The quotient of the number of entries in the first

n

rows which are not

1

’s and the number of

1

’s is:

\dfrac{n^2-n}{2n-1}

\dfrac{n^2-n}{4n-2}

\dfrac{n^2-2n}{2n-1}

\dfrac{n^2-3n+2}{4n-2}

None of these

Count all entries and then subtract the boundary

1

’s

The first

n

rows contain

\frac{n(n+1)}{2}

entries and

2n-1

boundary

1

’s

The first

n

rows contain

1+2+\cdots+n=\frac{n(n+1)}{2}

entries. There are

2n-1

boundary

1

’s, so the number of other entries is

\begin{gathered} \frac{n(n+1)}2-(2n-1)\\ =\frac{n^2+n-4n+2}{2}\\ =\frac{n^2-3n+2}{2}. \end{gathered}

Dividing by

2n-1

gives

\frac{n^2-3n+2}{4n-2}.

Therefore, the correct answer is D.

Problem 24 in other years