Question

fie A=7/7·9+7/9·11+7/11·13+....+7/(n-2).Determinati numarul n apartine N* astfel incat A=2001/4016

Answer 1

7* ($\frac{1}{7*9}+ \frac{1}{9*11} +......+ \frac{1}{(n-4)*(n-2)} + \frac{1}{(n-2)*n}$
Inmultesc cu 2/2 expresia 

=[tex] \frac{7}{2}*( \frac{9-7}{7*9}+....+ \frac{n-n+2}{n*(n-2)} )= [/tex]

Apoi se desparte 9*7/7*9 in 9/7*9 - 7/7*9 = 1/7- 1/9 ....si n-(n-2)/n*(n-2) in
 n/n*(n-2) - (n-2)/n*(n-2)= 1/n-2 - 1/n
Cand se scriu toate aceste fractii se reduc si nu mai ramane in paranteza decat prima si ultima adica 1/7-1/n
Expresia devine

A=$\frac{7}{2} * ( \frac{1}{7} - \frac{1}{n} )= \frac{2001}{4016}$
Rezulta ca [tex] \frac{7}{2}*( \frac{1}{7} - \frac{1}{n} = \frac{2001}{4016} [/tex]
Rezolvand aceasta ecuatie dupa ce aduci la acelasi numitor vei obtine
[tex] \frac{n-7}{2n} = \frac{2001}{4016} [/tex]

Inmultesc cu 2  in stg si dreapta si fac simplificarea prin 2 si obtin

$\frac{n-7}{n} = \frac{2001}{2008}$

fac produsul mezilor= produsul extremilor
2001n=2008n-2008*7
7n=2008*7
n=2008