Difference between revisions of "1971 AHSME Problems/Problem 24"
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== Solution == | == Solution == | ||
− | <math>\boxed{\textbf{(D) }\frac{n^2-3n+2}{4n-2}}</math>. | + | The first row has <math>1</math> one, and all the other rows have <math>2</math> ones. Thus, the total number of ones in the first <math>n</math> rows is <math>1+2(n-1)=2n-1</math>. The total number of elements in the first <math>n</math> rows is <math>1+2+3+\ldots+n</math>, which is the <math>n</math>th [[triangular number]], <math>\tfrac{n(n+1)}2</math>. Thus, the number of elements excluding ones is <math>\tfrac{n(n+1)}2-(2n-1)</math>. Thus, our desired quotient is <math>\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}}</math>. |
== See Also == | == See Also == | ||
{{AHSME 35p box|year=1971|num-b=23|num-a=25}} | {{AHSME 35p box|year=1971|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 17:47, 6 August 2024
Problem
Solution
The first row has one, and all the other rows have ones. Thus, the total number of ones in the first rows is . The total number of elements in the first rows is , which is the th triangular number, . Thus, the number of elements excluding ones is . Thus, our desired quotient is .
See Also
1971 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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All AHSME Problems and Solutions |
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