Difference between revisions of "1996 AHSME Problems/Problem 24"

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==Problem 24==
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==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==
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The sum of the first <math>1</math> numbers is <math>1</math>
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The sum of the next <math>2</math> numbers is <math>2 + 1</math>
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The sum of the next <math>3</math> numbers is <math>2 + 2 + 1</math>
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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.
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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>
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<math>f(49) = 1225</math>
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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>.
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We want to find:
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<math>\sum_{n=1}^{49} 1 + 2(n-1)</math>
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<math>\sum_{n=1}^{49} 2n - 1</math>
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<math>\sum_{n=1}^{49} 2n - \sum_{n=1}^{49} 1</math>
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<math>2 \sum_{n=1}^{49} n - 49</math>
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<math>2\cdot \frac{49\cdot 50}{2} - 49</math>
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<math>49^2</math>
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<math>2401</math>
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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>.
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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.
  
 
==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 14:44, 20 August 2011

Problem

The sequence $1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2,\ldots$ consists of $1$’s separated by blocks of $2$’s with $n$ $2$’s in the $n^{th}$ block. The sum of the first $1234$ terms of this sequence is

$\text{(A)}\ 1996\qquad\text{(B)}\ 2419\qquad\text{(C)}\ 2429\qquad\text{(D)}\ 2439\qquad\text{(E)}\ 2449$

Solution

The sum of the first $1$ numbers is $1$

The sum of the next $2$ numbers is $2 + 1$

The sum of the next $3$ numbers is $2 + 2 + 1$

In genereal, we can write "the sum of the next $n$ numbers is $1 + 2(n-1)$", where the word "next" follows the pattern established above.

Thus, we first want to find what triangular numbers $1234$ is between. By plugging in various values of $n$ into $f(n) = \frac{n(n+1)}{2}$, we find:

$f(50) = 1275$

$f(49) = 1225$

Thus, we want to add up all those sums from "next $1$ number" to the "next $49$ numbers", which will give us all the numbers up to and including the $1225^{th}$ number. Then, we can manually tack on the remaining $2$s to hit $1234$.

We want to find:

$\sum_{n=1}^{49} 1 + 2(n-1)$

$\sum_{n=1}^{49} 2n - 1$

$\sum_{n=1}^{49} 2n - \sum_{n=1}^{49} 1$

$2 \sum_{n=1}^{49} n - 49$

$2\cdot \frac{49\cdot 50}{2} - 49$

$49^2$

$2401$

Thus, the sum of the first $1225$ terms is $2401$. We have to add $9$ more $2$s to get to the $1234^{th}$ term, which gives us $2419$, or option $\boxed{B}$.

Note: If you notice that the above sums form $1 + 3 + 5 + 7... + (2n-1) = n^2$, the fact that $49^2$ appears at the end should come as no surprise.

See also

1996 AHSME (ProblemsAnswer KeyResources)
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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