Question

se se determine numărul natural x din egalitatea 1+5+9+...+x=231

Answer 1

$1+5+9+...+x=231 \\ \\ (4\cdot 1-3) +(4\cdot 2-3) + ... + \Big(4\cdot \dfrac{x+3}{4} - 3\Big) = 231 \\ \\ \boxed{4\cdot y -3 = x \Rightarrow 4y = x+3 \Rightarrow y = \dfrac{x+3}{4}} \\ \\ 4\cdot \Big(1+2+3+...+\dfrac{x+3}{4}\Big) +\\ +(-3-3-\underset{de~\frac{x+3}{4}~ ori}{\underbrace{...}}-3) = 231 \\\\ \\ 4\cdot \dfrac{\dfrac{x+3}{4}\cdot \Big(\dfrac{x+3}{4} +1 \Big)}{2} - 3\cdot \dfrac{x+3}{4} = 231 \\ \\ 2 \cdot \dfrac{x+3}{4} \cdot \Big(\dfrac{x+3}{4} + 1 \Big) - 3 \cdot \dfrac{x+3}{4} = 231\Big| \cdot 4 \cdot 4\\ \\ 2(x+3)(x+3+4)-12(x+3) = 231 \cdot 16\\ \\ 2(x+3)(x+7)-12(x+3) = 231 \cdot 16 \\ \\ (x+3)(2x+14-12) = 231\cdot 16\\ \\ 2(x+1)(x+3) = 231 \cdot 16 \\ \\ (x+1)(x+3) = 231 \cdot 8 \\ \\ (x+1)(x+3) = 11\cdot 21 \cdot 8 \\ \\ (x+1)(x+3) = 11 \cdot 3 \cdot 7 \cdot 2\cdot 4\\ \\ (x+1)(x+3) = 42 \cdot 44 \\ \\ \boxed{\boxed{x = 41}}$

Answer 2

   
[tex]\displaystyle\\ \text{Aplicam Formula lui Gauss:}\\\\ S = \frac{n(\text{ultimul termen }+\text{ primul termen})}{2}\\ \text{unde n = numarul de termeni din sir.}\\\\ n=\boxed{\frac{x-1}{4}+1}\\\\ S = \frac{n(x+1)}{2}=\frac{\left(\dfrac{x-1}{4}+1\right)\!\!\Big(x+1\Big)}{2} = 231 \\\\ \frac{\left(\dfrac{x-1}{4}+1\right)\!\!\Big(x+1\Big)}{2}=231\\\\ \left(\dfrac{x-1}{4}+\frac{4}{4}\right)\!\!\Big(x+1\Big)=231\times2\\\\ \left(\dfrac{x-1+4}{4}\right)\!\!\Big(x+1\Big)=462[/tex]


[tex]\displaystyle\\ \left(\dfrac{x+3}{4}\right)\Big(x+1\Big)=462\\\\ \frac{(x+3)(x+1)}{4}=462\\\\ (x+3)(x+1)=462\times4\\\\ x^2+4x+3=1848\\ x^2+4x+3-1848=0\\ x^2+4x-1845=0\\ x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-4\pm\sqrt{16+4\times1845}}{2}= \\ =\frac{-4\pm\sqrt{16+7380}}{2}=\frac{-4\pm\sqrt{7396}}{2}=\frac{-4\pm86}{2} \\ x_1 = \frac{-4+86}{2}=\frac{82}{2}=\boxed{\bf41}\\ x_2 = \frac{-4-86}{2}=\frac{-90}{2}=-45~\notin N~\text{Se respinge.} \\\\ \text{Solutie unica: }\boxed{\bf x=41} [/tex]