Question

Mă poate ajuta cineva, va rog?

Answer 1

$\displaystyle \sum\limits_{k=1}^{n}\Big[3(k-1)\Big]+4+\sum\limits_{k=1}^{n}\dfrac{8+3k}{2}= 137\\ \\ \sum\limits_{k=1}^{n}\Bigg[3(k-1)+\dfrac{8+3k}{2}\Bigg] = 133 \\ \\ \sum\limits_{k=1}^{n}\dfrac{9k+2}{2}= 133$

$\\ \\ \sum\limits_{k=1}^{n}\Big(\dfrac{9k}{2}+1\Big) = 133\\ \\ \dfrac{9}{2}\sum\limits_{k=1}^{n}(k)+n = 133 \\ \\ \dfrac{9}{2}\sum\limits_{k=1}^{n}k = 133-n \\ \\ \\\dfrac{9n(n+1)}{4} = 133-n\\ \\ 9n(n+1) = 532-4n \\ \\ 9n^2+13n-532 = 0\\ \\ \Delta = 19321 = 139^2 \\ \\ n = \dfrac{-13+139}{18} \Rightarrow \boxed{n = 7}$