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Sa se afle numărul natural x dacă 1 +2+3+...+x=210

Răspuns :

SUMA LUI GAUSS:

[tex]\boxed{1+2+3+...+n=\frac{n(n+1)}{2}}[/tex]

DEMONSTRATIE:

S=1  + 2   +   3   +......+n

S=n+(n-1)+(n-2)+.......+1

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2S=(n+1)+(n-1+2)+(n-2+3)+....+(1+n)

2S=(n+1)+(n+1)+....+(n+1)

(de n ori (n+1))

2S=n(n+1)

S=n(n+1):2

Explicație pas cu pas:

[tex]1+2+3+...+x=\frac{x(x+1)}{2}\\ \\ \frac{x(x+1)}{2}=210 \Rightarrow x(x+1)=420\\ \\ \Rightarrow x=20;(20\cdot21=420)\\ \\ SAU\\ \\ x^2+x-420=0\\ \\ \Delta=1+4\cdot420=1680 \Rightarrow x1=\frac{-1+41}{2}=20\\ cealalta..radacina..e..negativa..nu..o..luam..in..considerare[/tex]

[tex]210 =1+2+3+...+x\\ 210 =x+(x-1)+(x-2)+...+2+1\\ \\ \text{Adunam cele 2 egalitati:}\\ \\420 = (x+1)+(x+1)+(x+1)+\underset{\text{de } x \text{ ori}}{\underbrace{...}}+(x+1) \\ \\ 420 = (x+1)\cdot x \\ \\ 42\cdot 10 = (x+1)\cdot x \\ \\ 21\cdot 20 = (x+1)\cdot x \\ \\ \Rightarrow \boxed{x = 20}[/tex]