1971 AHSME Problems/Problem 24

Problem

$[asy] label("$1$",(0,0),S); label("$1$",(-1,-1),S); label("$1$",(-2,-2),S); label("$1$",(-3,-3),S); label("$1$",(-4,-4),S); label("$1$",(1,-1),S); label("$1$",(2,-2),S); label("$1$",(3,-3),S); label("$1$",(4,-4),S); label("$2$",(0,-2),S); label("$3$",(-1,-3),S); label("$3$",(1,-3),S); label("$4$",(-2,-4),S); label("$4$",(2,-4),S); label("$6$",(0,-4),S); label("etc.",(0,-5),S); //Credit to chezbgone2 for the diagram[/asy]$

Solution

The first row has $1$ one, and all the other rows have $2$ ones. Thus, the total number of ones in the first $n$ rows is $1+2(n-1)=2n-1$ . The total number of elements in the first $n$ rows is $1+2+3+\ldots+n$ , which is the $n$ th triangular number, $\tfrac{n(n+1)}2$ . Thus, the number of elements excluding ones is $\tfrac{n(n+1)}2-(2n-1)$ . Thus, our desired quotient is $\frac{\tfrac{n(n+1)}2-(2n-1)}{2n-1} = \frac{n^2+n-2(2n-1)}{2(2n-1)} = \boxed{\textbf{(D) }\frac{n^2-3n+2}{4n-2}}$ .

1971 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
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1971 AHSME Problems/Problem 24

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