Question

Aratati ca numarul B=1+3^1+3^2+.......+3^61 este divizibil cu 4

Answer 1

    
[tex]B=1+3^1+3^2+3^3+...+3^{60}+3^{61} ~~~~unde: ~~~ 1=3^0 \\ n = \text{numarul de termeni } = 61+1=\boxed{62} \\ \text{Observam ca suma primilor 2 termeni = } 1 + 3 = \boxed{4} \\ Vom grupa sirul de termeni in grupe de cate 2 termeni. \\ \text{Avem voie deoarece n = 62 de termeni, si 62 este divizibil cu 2.} \\ B=(1+3^1)+(3^2+3^3)+(3^4+3^5)+...+(3^{60}+3^{61})= \\ = 1(1+3^1)+3^2(1+3^1)+3^4(1+3^1)+...+3^{60}(1+3^1)= \\=(1+3^1)(1+3^2+3^4+......+3^{60}) =\boxed{4(1+3^2+3^4+......+3^{60})\;\vdots\;4}[/tex]