Question

Aflati nr real x astfel incat 1+5+9+...+x=496

Answer 1

   
[tex]\displaystyle\\ 1+5+9+...+x=496\\ \text{Adunam termenii sirului cu formula lui Gauss.}\\\\ S = \frac{n(x+1)}{2}~~~\text{unde n este numarul de termeni din sir.}\\\\ \text{Calculam numarul de termeni:}\\\\ n=\frac{x-1}{4}+1 = \frac{x-1}{4}+\frac{4}{4} =\frac{x-1+4}{4}=\boxed{\bf \frac{x+3}{4}~termeni}\\\\ \text{Calculam suma sirului:}\\\\\\ S = \frac{ \dfrac{x+3}{4} \cdot (x+1)}{2}= \frac{(x+3)(x+1)}{8}\\\\ [/tex]


[tex]\displaystyle\\ \text{Scriem ecuatia:}\\\\ \frac{(x+3)(x+1)}{8}=496\\\\ (x+3)(x+1)=496\cdot 8\\\\ x^2+4x+3 = 3968\\\\ x^2+4x+3 - 3968 = 0\\\\ x^2+4x - 3965 = 0\\\\ x_{12}= \frac{-b\pm \sqrt{b^2-4ac} }{2a}=\frac{-4\pm\sqrt{16+4\cdot 3965} }{2}=\\\\ =\frac{-4\pm\sqrt{16+15860} }{2}=\frac{-4\pm\sqrt{15876} }{2}= \frac{=-4\pm 126}{2} = -2\pm 63\\\\ x_1 = -2+63 = \boxed{\bf 61}\\\\ x_2 = -2-63 = -65 ~~~-65 \ \textless \ 0;~~-65 \notin N \\\\ \Longrightarrow~~\boxed{\bf x= 61}[/tex]

Answer 2

$\it 1 + 5 + 9 + ... + x = 496$


Membrul stâng reprezintă suma termenilor unei progresii aritmetice

cu rația 4.

[tex]\it x = a_n=a_1+(n-1)r =1+(n-1)\cdot4 = 4n - 3 \ \ \ \ (1) \\ \\ \\ S_n = \dfrac{(a_1+a_n)\cdot n}{2} = 496 \Rightarrow \dfrac{(1+4n-3)\cdot n}{2} =496 \Rightarrow \\ \\ \\ \Rightarrow \dfrac{(4n-2)\cdot n}{2} = 496\Rightarrow \dfrac{2(2n-1)\cdot n}{2} = 496 \Rightarrow (2n-1) \cdot n = 496 \Rightarrow \\ \\ \\ \Rightarrow n(2n-1) = 496 \Rightarrow n(2n-1) = 16\cdot 31 \Rightarrow n = 16 \ \ \ \ \ (2) [/tex]

$\it (1), (2) \Rightarrow x = 4\cdot16-3 = 61$