AMC 12 Mock Exam 4 Problem 11

Compute: $a_3=\frac{3+1}{2}=2$, $a_4=\frac{2+1}{3}=1$, $a_5=\frac{1+1}{2}=1$, $a_6=\frac{1+1}{1}=2$, $a_7=\frac{2+1}{1}=3$. Since $a_6=a_1$ and $a_7=a_2$, and each term depends only on the two before it, the sequence repeats with period $5$: $2,3,2,1,1,2,3,2,1,1,\dots$. One period sums to $9$. As $2026=5\cdot405+1$, the sum is $405\cdot9+a_{2026}=3645+a_1=3645+2=\boxed{3647}$.