Difference between revisions of "2002 AMC 10B Problems/Problem 23"

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Problem 23

Let $\{a_k\}$ be a sequence of integers such that $a_1=1$ and $a_{m+n}=a_m+a_n+mn,$ for all positive integers $m$ and $n.$ Then $a_{12}$ is

$\mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 56\qquad \mathrm{(C) \ } 67\qquad \mathrm{(D) \ } 78\qquad \mathrm{(E) \ } 89$

Solution 1

When $m=1$ , $a_{n+1}=1+a_n+n$ . Hence, $a_{2}=1+a_1+2$ $a_{3}=1+a_2+3$ $a_{4}=1+a_3+4$ $\dots$ $a_{12}=1+a_{11}+11$ Adding these equations up, we have that $a_12=12+(1+2+3+...+11)=\boxed{78}$

Solution 2

Substituting $n=1$ into $a_{m+n}=a_m+a_n+mn$ : $a_{m+1}=a_m+a_{1}+m$ . Since $a_1 = 1$ , $a_{m+1}=a_m+m+1$ . Therefore, $a_m = a_{m-1} + m, a_{m-1}=a_{m-2}+(m-1), a_{m-2} = a_{m-3} + (m-2)$ , and so on until $a_2 = a_1 + 2$ . Adding the Left Hand Sides of all of these equations gives $a_m + a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_2$ ; adding the Right Hand Sides of these equations gives $(a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_1) + (m + (m-1) + (m-2) + \cdots + 2)$ . These two expressions must be equal; hence $a_m + a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_2 = (a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_1) + (m + (m-1) + (m-2) + \cdots + 2)$ and $a_m = a_1 + (m + (m-1) + (m-2) + \cdots + 2)$ . Substituting $a_1 = 1$ : $a_m = 1 + (m + (m-1) + (m-2) + \cdots + 2) = 1+2+3+4+ \cdots +m = \frac{(m+1)(m)}{2}$ . Thus we have a general formula for $a_m$ and substituting $m=12$ : $a_{12} = \frac{(13)(12)}{2} = (13)(6) = \boxed{\mathrm{(D) \ } 78}$ .

Solution 3

We can literally just plug stuff in. No prerequisite is actually said in the sequence. Since $a_{m+n} = a_m+a_n +mn$ , we know $a_2=a_1+a_1+1\cdot1=1+1+1=3$ . After this, we can use $a_2$ to find $a_4$ . $a_4=a_2+a_2+2\cdot 2 = 3+3+4 = 10$ . Now, we can use $a_2$ and $a_4$ to find $a_6$ , or $a_6=a_4+a_2+4\cdot 2 = 10+3+8=21$ . Lastly, we can use $a_6$ to find $a_{12}$ . $a_{12} = a_6+a_6+6\cdot 6 = 21+21+36= \boxed{(D) 78}$

Additional Comment

This is also the formula for the triangular numbers $T_n=1+2+3+...+n$ , as seen in Solution 2

@@ Line 7: / Line 7: @@
 == Solution 1 ==
-First of all, write <math>a_3</math> and <math>a_4</math> in terms of <math>a_2.</math>
+When <math>m=1</math>, <math>a_{n+1}=1+a_n+n</math>. Hence,
-<cmath>\begin{align*} a_3 &= a_{1+2} = 1 + a_2 + 2 = a_2 + 3\\
+<cmath>a_{2}=1+a_1+2</cmath>
-a_4 &= a_{2+2} = a_2 + a_2 + 4 = 2a_2 + 4 \end{align*}</cmath>
+<cmath>a_{3}=1+a_2+3</cmath>
+<cmath>a_{4}=1+a_3+4</cmath>
-<math>a_6</math> can be represented by <math>a_2</math> in <math>2</math> different ways.
+<math>\dots</math>
+<cmath>a_{12}=1+a_{11}+11</cmath>
-<cmath>\begin{align*} a_{2+4} &= a_2 + a_4 + 8 = a_2 + 2a_2 + 4 + 8 = 3a_2 + 12\\
+Adding these equations up, we have that <math>a_12=12+(1+2+3+...+11)=\boxed{78}</math>
-a_{3+3} &= 2a_3 + 9 = 6 + 2a_2 + 9 = 2a_2 + 15\\
-\end{align*}</cmath>
-Since both are equal to <math>a_6,</math> you can set them equal to each other.
-<cmath>\begin{align*} 3a_2 + 12 &= 2a_2 + 15\\
-a_2 &= 3 \end{align*}</cmath>
-Substitute the value of <math>a_2</math> back into <math>a_6,</math> and substitute that into <math>a_{12}.</math>
-<cmath>a_6 = 2a_2+15 = 6+15 = 21</cmath>
-<cmath>a_{12} = a_{6+6} = a_6+a_6+36 = 21+21+36 = \boxed{\mathrm{(D) \ } 78}</cmath>
 == Solution 2 ==

2002 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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