Question

Determine the second smallest prime factor of $\frac{1^3+1}{1+1}+\frac{2^3+1}{2+1}+\frac{3^3+1}{3+1}+\cdots +\frac{{2005}^3+1}{2005+1}.$
(correct answer + 3, wrong answer 0)

Answer 1

Let $S=\frac{1^3+1}{1+1}+\frac{2^3+1}{2+1}+\frac{3^3+1}{3+1}+\cdots +\frac{{2005}^3+1}{2005+1}$
$\Rightarrow S=\sum _{r=1}^{2005}(r^2-r+1)\Rightarrow S=\frac{1}{6}(2005\times 2006\times 4011)-\frac{1}{2}(2005\times 2006)+2005\Rightarrow S=5\times 401\times 1340009\Rightarrow S=5\times 11\times 401\times 121819$

Since $2,3,7$ does not divide $401$ and $121819,$ the second smallest prime factor is $11.$