Difference between revisions of "1996 AHSME Problems/Problem 24"
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| − | ==Problem | + | ==Problem== |
The sequence <math> 1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2,\ldots </math> consists of <math>1</math>’s separated by blocks of <math>2</math>’s with <math>n</math> <math>2</math>’s in the <math>n^{th}</math> block. The sum of the first <math>1234</math> terms of this sequence is | The sequence <math> 1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2,\ldots </math> consists of <math>1</math>’s separated by blocks of <math>2</math>’s with <math>n</math> <math>2</math>’s in the <math>n^{th}</math> block. The sum of the first <math>1234</math> terms of this sequence is | ||
<math> \text{(A)}\ 1996\qquad\text{(B)}\ 2419\qquad\text{(C)}\ 2429\qquad\text{(D)}\ 2439\qquad\text{(E)}\ 2449 </math> | <math> \text{(A)}\ 1996\qquad\text{(B)}\ 2419\qquad\text{(C)}\ 2429\qquad\text{(D)}\ 2439\qquad\text{(E)}\ 2449 </math> | ||
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| + | ==Solution== | ||
| + | |||
| + | The sum of the first <math>1</math> numbers is <math>1</math> | ||
| + | |||
| + | The sum of the next <math>2</math> numbers is <math>2 + 1</math> | ||
| + | |||
| + | The sum of the next <math>3</math> numbers is <math>2 + 2 + 1</math> | ||
| + | |||
| + | In genereal, we can write "the sum of the next <math>n</math> numbers is <math>1 + 2(n-1)</math>", where the word "next" follows the pattern established above. | ||
| + | |||
| + | Thus, we first want to find what triangular numbers <math>1234</math> is between. By plugging in various values of <math>n</math> into <math>f(n) = \frac{n(n+1)}{2}</math>, we find: | ||
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| + | <math>f(50) = 1275</math> | ||
| + | |||
| + | <math>f(49) = 1225</math> | ||
| + | |||
| + | Thus, we want to add up all those sums from "next <math>1</math> number" to the "next <math>49</math> numbers", which will give us all the numbers up to and including the <math>1225^{th}</math> number. Then, we can manually tack on the remaining <math>2</math>s to hit <math>1234</math>. | ||
| + | |||
| + | We want to find: | ||
| + | |||
| + | <math>\sum_{n=1}^{49} 1 + 2(n-1)</math> | ||
| + | |||
| + | <math>\sum_{n=1}^{49} 2n - 1</math> | ||
| + | |||
| + | <math>\sum_{n=1}^{49} 2n - \sum_{n=1}^{49} 1</math> | ||
| + | |||
| + | <math>2 \sum_{n=1}^{49} n - 49</math> | ||
| + | |||
| + | <math>2\cdot \frac{49\cdot 50}{2} - 49</math> | ||
| + | |||
| + | <math>49^2</math> | ||
| + | |||
| + | <math>2401</math> | ||
| + | |||
| + | Thus, the sum of the first <math>1225</math> terms is <math>2401</math>. We have to add <math>9</math> more <math>2</math>s to get to the <math>1234^{th}</math> term, which gives us <math>2419</math>, or option <math>\boxed{B}</math>. | ||
| + | |||
| + | Note: If you notice that the above sums form <math>1 + 3 + 5 + 7... + (2n-1) = n^2</math>, the fact that <math>49^2</math> appears at the end should come as no surprise. | ||
==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=23|num-a=25}} | {{AHSME box|year=1996|num-b=23|num-a=25}} | ||
Revision as of 15:44, 20 August 2011
Problem
The sequence 
consists of 
’s separated by blocks of 
’s with 
’s in the 
block. The sum of the first 
terms of this sequence is
Solution
The sum of the first numbers is
The sum of the next numbers is 
The sum of the next 
numbers is 
In genereal, we can write "the sum of the next numbers is 
", where the word "next" follows the pattern established above.
Thus, we first want to find what triangular numbers is between. By plugging in various values of into 
, we find:
Thus, we want to add up all those sums from "next number" to the "next 
numbers", which will give us all the numbers up to and including the 
number. Then, we can manually tack on the remaining s to hit .
We want to find:
Thus, the sum of the first 
terms is . We have to add 
more s to get to the 
term, which gives us 
, or option 
.
Note: If you notice that the above sums form 
, the fact that appears at the end should come as no surprise.
See also
| 1996 AHSME (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 | ||
| All AHSME Problems and Solutions | ||
