Folosim relatia:[tex] \frac{1}{n-2}-\frac{1}{n}=\frac{2}{(n-2)n} [/tex]
Atunci relatia initiala devine:
[tex] \frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7} +...\frac{1}{n-2}-\frac{1}{n}=\frac{332}{999} [/tex]
Dupa reducerea termenilor asemenea se obtine:
[tex] \frac{1}{3}-\frac{1}{n}=\frac{332}{999} [/tex]
Aducem la acelasi numitor in membrul stang:
[tex] \frac{n-3}{3n} =\frac{332}{999} [/tex]⇒999(n-3)=332*3n⇒999n-2997=996n⇒999n-996n=2997⇒3n=2997⇒n=999
