Question

Fie A= (1/1*3+1/3*5+1/5*7+...+1/2013*2015): (2014/2015)*3 la n+1,
a)Aratati ca A=3 la 2n+1
b)determinati n apartine N*, astfel incat A= 3 la 4n+5

Answer 1

Daca notez cu S prima paranteza a lui A, atunci avem:

$2S=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2013\cdot2015}=$

$=\dfrac{3-1}{1\cdot3}+\dfrac{5-3}{3\cdot5}+\dfrac{7-5}{5\cdot7}+...+\dfrac{2015-2013}{2013\cdot2015}=$

$=\dfrac{3}{1\cdot3}-\dfrac{1}{1\cdot3}+\dfrac{5}{3\cdot5}-\dfrac{3}{3\cdot5}+\dfrac{7}{5\cdot7}-\dfrac{5}{5\cdot7}+...$

$...+\dfrac{2015}{2013\cdot2015}-\dfrac{2013}{2013\cdot2015}=$

Dupa simplificari

$=\dfrac11-\dfrac13+\dfrac13-\dfrac15-\dfrac15+\dfrac17-\dfrac17+...+\dfrac{1}{2013}-\dfrac{1}{2015}=$

dupa ce reducem termenii asemenea

$=1-\dfrac{1}{2015}=\dfrac{2014}{2015}$

In continuare este simplu de constatat ca

$A=\dfrac{3^{n+1}}{2}$

Metoda aceasta de calcul a sumei se poate aplica la foarte multe sume de acest gen.