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
a2a3a6=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