Question

(1/2+2/3+3/4....+99/100) ×(17/51-19/57)=​

Answer 1

$\frac{1}{2}=1-\frac{1}{2}$

$\frac{2}{3} =1-\frac{1}{3}$

$\frac{3}{4} =1-\frac{1}{4}$

...

$\frac{99}{100}=1-\frac{1}{100}$

Notam:

$S=\frac{1}{2}+\frac{2}{3} +\frac{3}{4}+...+\frac{99}{100}$

$S=1-\frac{1}{2}+1-\frac{1}{3}+1-\frac{1}{4}+...+1-\frac{1}{100}$

Avem 99 de 1

$S=99-(\frac{1}{2} +\frac{1}{3} +...+\frac{1}{100})$

Calculam $x=\frac{1}{2} +\frac{1}{3}+...+\frac{1}{100}$

Stim ca:

$\frac{1}{(n+1)(n+2)} =\frac{1}{n+1} -\frac{1}{n+2}$

$\[ \sum_{i=1}^{99} \frac{1}{n+1} \]$=$\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}$

$\[ \sum_{i=1}^{99} \frac{1}{n+2} \]$=$\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}+\frac{1}{101}$

$\[ \sum_{i=1}^{99} \frac{1}{(n+1)(n+2)} \]=$$\[ \sum_{i=1}^{99} \frac{1}{n+1} \]-$$\[ \sum_{i=1}^{99} \frac{1}{n+2} \]$

$\[ \sum_{i=1}^{99} \frac{1}{(n+1)(n+2)} \]=$$\frac{1}{2} -\frac{1}{101} =\frac{99}{202}$

$x=\frac{99}{202}$

$S=99-\frac{99}{202} =\frac{99\times201}{202}$

Deci rezultatul nostru va fi:

$\frac{99\times201}{202}\times(\frac{17}{51}-\frac{19}{57} )=\\\\\frac{99\times201}{202}\times (\frac{1}{3}-\frac{1}{3})= \frac{99\times201}{202}\times 0=0$