Question

Să se arate ca : subpunctul b) va rog

Answer 1

$\arctan(x)-\arctan(y) = \arctan\dfrac{x-y}{1+xy} \\ \\ \\\arctan\dfrac{1}{k^2+k+1} = \arctan\dfrac{(k+1)-k}{1+(k+1)k} = \arctan(k+1)-\arctan(k) \\ \\ \\\displaystyle \sum\limits_{k=1}^n\arctan\dfrac{1}{k^2+k+1} = \sum\limits_{k=1}^n\Big[ \arctan(k+1)-\arctan(k)\Big] = \\ \\ = \arctan 2+\arctan 3+...+\arctan n +\arctan(n+1) - \\ -\arctan 1-\arctan 2-\arctan 3-...-\arctan n = \\ \\ =\arctan(n+1)-\arctan 1 = \arctan \dfrac{(n+1)-1}{1+(n+1)\cdot 1} = \boxed{\arctan \dfrac{n}{n+2}}$