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 $a_{12}=12+(1+2+3+...+11)=\boxed{78}$

~AopsUser101

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 $a_{12}$ . $a_{12} = a_6+a_6+6\cdot 6 = 21+21+36= \boxed{(D) 78}$

Solution 4

We can set $n$ equal to $m$ , so we can say that $a_{m + m} = a_m + a_m = m*m$ $a_{2m} = 2a_m + m^2$

We set $2m = 12$ , we get $m = 6$ . $a_{12} = 2a_6 + 36$

We set $2m = 6$ m, we get $m = 3$ . $a_6 = 2a_3 + 9$

Solving for $a_3$ is easy, just direct substitution. $a_2 = 1 + 1 + 1 = 3$ $a_3 = a_{2 + 1} = 3 + 1 + 2 = 6$

Substituting, we get $a_6 = 2(6) + 9 = 21$ $a_12 = 2(21) + 36 = 78$

Thus, the answer is $\boxed{D}$ .

Additional Comment

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

@@ Line 26: / Line 26: @@
 ==Solution 4==
+We can set <math>n</math> equal to <math>m</math>, so we can say that
+<cmath>a_{m + m} = a_m + a_m = m*m</cmath>
+<cmath>a_{2m} = 2a_m + m^2</cmath>
+We set <math>2m = 12</math>, we get <math>m = 6</math>.
+<cmath>a_{12} = 2a_6 + 36</cmath>
+We set <math>2m = 6</math>m, we get <math>m = 3</math>.
+<cmath>a_6 = 2a_3 + 9</cmath>
+Solving for <math>a_3</math> is easy, just direct substitution.
+<cmath>a_2 = 1 + 1 + 1 = 3</cmath>
+<cmath>a_3 = a_{2 + 1} = 3 + 1 + 2 = 6</cmath>
+Substituting, we get
+<cmath>a_6 = 2(6) + 9 = 21</cmath>
+<cmath>a_12 = 2(21) + 36 = 78</cmath>
+Thus, the answer is <math>\boxed{D}</math>.
 == Additional Comment==

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