Răspuns :
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
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
[tex]\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} [/tex]