Difference between revisions of "2002 AMC 12P Problems/Problem 11"

Revision as of 00:45, 31 December 2023

Problem

Let $t_n = \frac{n(n+1)}{2}$ be the $n$ th triangular number. Find

$\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} + ... + \frac{1}{t_{2002}}$

$\text{(A) }\frac {4003}{2003} \qquad \text{(B) }\frac {2001}{1001} \qquad \text{(C) }\frac {4004}{2003} \qquad \text{(D) }\frac {4001}{2001} \qquad \text{(E) }2$

Solution

If $\log_{b} 729 = n$ , then $b^n = 729$ . Since $729 = 3^6$ , $b$ must be $3$ to some factor of 6. Thus, there are four (3, 9, 27, 729) possible values of $b \Longrightarrow \boxed{\mathrm{E}}$ .

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 2: / Line 2: @@
 Let <math>t_n = \frac{n(n+1)}{2}</math> be the <math>n</math>th triangular number. Find
-<cmath>\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} + ... + \frac{1}{t_2002}</cmath>
+<cmath>\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} + ... + \frac{1}{t_{2002}}</cmath>
 <math>

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Difference between revisions of "2002 AMC 12P Problems/Problem 11"

Revision as of 00:45, 31 December 2023

Problem

Solution

See also