7* ([tex] \frac{1}{7*9}+ \frac{1}{9*11} +......+ \frac{1}{(n-4)*(n-2)} + \frac{1}{(n-2)*n} [/tex]
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=[tex] \frac{7}{2} * ( \frac{1}{7} - \frac{1}{n} )= \frac{2001}{4016} [/tex]
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
[tex] \frac{n-7}{n} = \frac{2001}{2008} [/tex]
fac produsul mezilor= produsul extremilor
2001n=2008n-2008*7
7n=2008*7
n=2008
