Question

Va rog, dau coroana!

Answer 1

1.          1 / n      - 1 / ( n +1 )  = ( n +1 - n ) / n·( n +1 ) = 1 / n·( n +1 ) 
                                   adevarat 
2. aplicam , pentru fiecare termen , formula 
1 / 1 · 1 =    1 / 1         - 1 /2 
1 / 2 ·  3 =  1  /2           - 1  /3 
 1 / 3  · 4 =  1 /3           -  1 / 4 
..........................................
1 / 2013 · 2014 = 1 / 2013  -   1 / 2014 
din adunare , raman doar primul si ultimul termen  ,ceilalti se reduc

suma       , ex.  = 1 /1   - 1 /2014  =( 2014 - 1 ) / 2014 = 2013  / 2014

Answer 2

$\displaystyle a). \frac{1}{n} - \frac{1}{n+m} = \frac{n+m}{n(n+m)} - \frac{n}{n(n+m)} = \frac{n+m-n}{n(n+m)} = \frac{m}{n(n+m)} \\ \\ b). \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} +...+ \frac{1}{2013 \cdot 2014} = \\ \\ = \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} +...+ \frac{1}{2013} - \frac{1}{2014} = \\ \\ = {^{2014)} 1- \frac{1}{2014} = \frac{2014-1}{2014} = \frac{2013}{2014}$