Question

x(1+1/2+1/3+......+1/2019)=2019-1/2-1/3-......-2018/2019

Answer 1

Răspuns:

1

Explicație pas cu pas:

Totul sta in ultimul sir. Daca dam factor comun -(minus), ramane cu toate pozitive:

$\frac{1}{2}+\frac{2}{3}+...+\frac{2018}{2019}=1-\frac{1}{2}+1-\frac{1}{3}+...+1-\frac{1}{2019}=2018-(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019})$

Daca nu intelegi ce s-a intamplat aici, explicatia e asa:

Sirul nostru e de forma $\frac{n}{n+1}$, care poate fi scris si ca $1-\frac{1}{n}$. Daca amplifici ai sa vezi ca da acelasi lucru. Acum ne intoarcem la exercitiul complet.

$x(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019})=2019-[2018-(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019})]\\x(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019})=2019-2018+(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019})\\x(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019})=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\\x=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}\\x=1$