Syler wrote: Determine strictly positive real numbers $ a_{1},a_{2},...,a_{n}$ if for any $ n\in N^*$ takes place equality: $ a_{1}^2 + a_{2}^2 + ... + a_{n}^2 = a_{1} + a_{2} + ... + a_{n} + \frac {n(n^2 + 6n + 11)}{3}$ So $ \sum_{k=1}^n(a_k^2-a_k)=\frac {n(n^2 + 6n + 11)}{3}$ and so $ \sum_{k=1}^{n-1}(a_k^2-a_k)=\frac {(n-1)((n-1)^2 + 6(n-1) + 11)}{3}$ Subtracting these two equalities implies $ a_n^2-a_n=n^2+3n+2$ $ \iff$ $ (a_n-n-2)(a_n+n+1)=0$ and so $ a_n=n+2$ (since $ a_n>0$), and it's immediate to check back that this indeed is a solution. Hence the answer : $ \boxed{a_n=n+2}$ And I wonder how you could know that the problem is nice if you dont know the solution

pco wrote: Syler wrote: Determine strictly positive real numbers $ a_{1},a_{2},...,a_{n}$ if for any $ n\in N^*$ takes place equality: $ a_{1}^2 + a_{2}^2 + ... + a_{n}^2 = a_{1} + a_{2} + ... + a_{n} + \frac {n(n^2 + 6n + 11)}{3}$ So $ \sum_{k = 1}^n(a_k^2 - a_k) = \frac {n(n^2 + 6n + 11)}{3}$ and so $ \sum_{k = 1}^{n - 1}(a_k^2 - a_k) = \frac {(n - 1)((n - 1)^2 + 6(n - 1) + 11)}{3}$ Subtracting these two equalities implies $ a_n^2 - a_n = n^2 + 3n + 2$ $ \iff$ $ (a_n - n - 2)(a_n + n + 1) = 0$ and so $ a_n = n + 2$ (since $ a_n > 0$), and it's immediate to check back that this indeed is a solution. Hence the answer : $ \boxed{a_n = n + 2}$ And I wonder how you could know that the problem is nice if you dont know the solution Thank you a lot. I supposed that the problem is nice.