Problem:
Let {ak​} be a sequence of integers such that a1​=1 and am+n​=am​+an​+mn, for all positive integers m and n. Then a12​ is
Answer Choices:
A. 45
B. 56
C. 67
D. 78
E. 89
Solution:
By setting n=1 in the given recursive equation, we obtain am+1​=am​+ a1​+m, for all positive integers m. So am+1​−am​=m+1 for each m=1,2,3,… Hence,
a12​−a11​=12,a11​−a10​=11,…,a2​−a1​=2.
Summing these equalities yields a12​−a1​=12+11+⋯+2. So
a12​=12+11+⋯+2+1=212(12+1)​=78.
OR
We have
a2​a3​a6​​=a1+1​=a1​+a1​+1⋅1=1+1+1=3,=a2+1​=a2​+a1​+2⋅1=3+1+2=6,=a3+3​=a3​+a3​+3⋅3=6+6+9=21,​
and
a12​=a6+6​=a6​+a6​+6⋅6=21+21+36=78.
Answer: D​.
The problems on this page are the property of the MAA's American Mathematics Competitions