35. Determine the second smallest prime factor of
$$
\frac{1^{3}+1}{1+1}+\frac{2^{3}+1}{2+1}+\frac{3^{3}+1}{3+1}+\cdots+\frac{2005^{3}+1}{2005+1}
$$
35 Ans: 11.
The sum $=\sum_{r=1}^{2005}\left(r^{2}-r+1\right)=\frac{1}{6} \times 2005 \times 2006 \times 4011-\frac{1}{2} \times 2005 \times 2006+2005$. Upon
simplifying, we see that the sum $=5 \times 401 \times 134009=5 \times 11 \times 401 \times 121819 \times$.
Since $2,3,7$ does not divide 401 and 121819 , the second smallest prime factor is 11 .