Difference between revisions of "1996 AHSME Problems/Problem 24"
XXQw3rtyXx (talk | contribs) (→Solution) |
|||
Line 42: | Line 42: | ||
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 2== | ||
+ | We note that the number of elements in each block are | ||
+ | |||
+ | <math>2, 3, 4,</math> and so on. | ||
+ | |||
+ | Let <math>n</math> be the number of blocks up to <math>1234</math> terms. | ||
+ | |||
+ | We have <math>\frac {2+n+1}{2} \cdot (n)=1234</math> | ||
+ | |||
+ | Solving for <math>n</math>, we get about <math>48.2</math>. | ||
+ | |||
+ | We have <math>48</math> full blocks and <math>1</math> partial block. We find that there are a total of <math>49</math> <math>1</math>'s | ||
+ | |||
+ | Now, we change every number in the sequence to <math>2</math>, and then add. We get <math>2468</math>. Since we added <math>49</math> by changing all <math>49</math> <math>1</math>'s to <math>2</math>, we must subract that <math>49</math>. Giving us | ||
+ | |||
+ | <math>2468-49=2419</math> <math>\boxed{B}</math> | ||
==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=23|num-a=25}} | {{AHSME box|year=1996|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:04, 21 September 2013
Contents
[hide]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.
Solution 2
We note that the number of elements in each block are
and so on.
Let be the number of blocks up to terms.
We have
Solving for , we get about .
We have full blocks and partial block. We find that there are a total of 's
Now, we change every number in the sequence to , and then add. We get . Since we added by changing all 's to , we must subract that . Giving us
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.