Browse · MATH Print → jmc algebra junior 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. Find a12. Solution — click to reveal We have that a2a3a6a12=a1+a1+1=3,=a1+a2+2=6,=a3+a3+9=21,=a6+a6+36=78. Final answer 78 ← Previous problem Next problem →