Difference between revisions of "1996 AHSME Problems/Problem 24"
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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. | 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. | ||
− | ==Solution | + | ==Solution 2== |
The <math>k</math>th appearance of 1 is at position <math>1 + 2 + \dots + k = \frac{k(k + 1)}{2}</math>. Then there are <math>k</math> 1's and <math>\frac{k(k + 1)}{2} - k = \frac{k(k - 1)}{2}</math> 2's among the first <math>\frac{k(k + 1)}{2}</math> numbers, so the sum of these <math>\frac{k(k + 1)}{2}</math> terms is <math>k + k(k - 1) = k^2</math>. | The <math>k</math>th appearance of 1 is at position <math>1 + 2 + \dots + k = \frac{k(k + 1)}{2}</math>. Then there are <math>k</math> 1's and <math>\frac{k(k + 1)}{2} - k = \frac{k(k - 1)}{2}</math> 2's among the first <math>\frac{k(k + 1)}{2}</math> numbers, so the sum of these <math>\frac{k(k + 1)}{2}</math> terms is <math>k + k(k - 1) = k^2</math>. | ||
Latest revision as of 13:02, 12 September 2020
Contents
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 general, 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.
Solution 2
The th appearance of 1 is at position . Then there are 1's and 2's among the first numbers, so the sum of these terms is .
When , , and when , .
The sum of the first 1225 terms is . The numbers in positions 1226 through 1234 are all 2's, so their sum is . Therefore, the sum of the first 1234 terms is .
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 |
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