# 2002 AMC 10B Problems/Problem 23

## 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, $$a_{2}=1+a_1+1$$ $$a_{3}=1+a_2+2$$ $$a_{4}=1+a_3+3$$ $$\dots$$ $$a_{12}=1+a_{11}+11$$ Adding these equations up, we have that $a_{12}=12+(1+2+3+...+11)=\boxed{\mathrm{(D) \ } 78}$

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

## 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 $\boxed{D}$.

~ euler123

## Solution 5

Note that the sequence of triangular numbers $T_n=1+2+3+...+n$ satisfies these conditions. It is immediately obvious that it satisfies $a_1=1$, and $a_{m+n}=a_m+a_n+mn$ can be visually proven with the diagram below.

$[asy] for(int i=5; i > 0; --i) { for(int j=0; j < i; ++j) { draw(circle((j+(5-i)/2,(5-i)*sqrt(3)/2),.2)); }; }; path m1 = brace((2,-.3),(0,-.3),.2); draw(m1); label("m",m1,S); path n1 = brace((4,-.3),(3,-.3),.2); draw(n1); label("n",n1,S); draw((-.2*sqrt(3),-.2)--(2+.2*sqrt(3),-.2)--(1,.4+sqrt(3))--cycle); label("T_m",(1,1/3*sqrt(3))); draw((3-.2*sqrt(3),-.2)--(4+.2*sqrt(3),-.2)--(3.5,.4+.5*sqrt(3))--cycle); label("T_n",(3.5,.5/3*sqrt(3))); path m2 = brace((2+.15*sqrt(3),.15+2*sqrt(3)),(3+.15*sqrt(3),.15+sqrt(3)),.2); draw(m2); label("m",m2,(.5*sqrt(3),.5)); path n2 = brace((1.5-.15*sqrt(3),.15+1.5*sqrt(3)),(2-.15*sqrt(3),.15+2*sqrt(3)),.2); draw(n2); label("n",n2,(-.5*sqrt(3),.5)); draw((2.5,-.4+.5*sqrt(3))--(3+.4/3*sqrt(3),sqrt(3))--(2,.4+2*sqrt(3))--(1.5-.4/3*sqrt(3),1.5*sqrt(3))--cycle); label("mn",(2.25,1.25*sqrt(3))); [/asy]$

This means that we can use the triangular number formula $T_n = \frac{n(n+1)}{2}$, so the answer is $T_{12} = \frac{12(12+1)}{2} = \boxed{\mathrm{(D) \ } 78}$. ~emerald_block

## See also

 2002 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 22 Followed byProblem 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 All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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