Acolo nu este 4, este n, pentru ca sumele sunt scrise in functie de n. Daca ar fi fost 4 ar fi fost simplu, pur si simplu calculai fiecare combinare in parte, nu ai fi avut ..., ai fi avut direct toti termenii sumei
a) In general avem urmatoarea ecuatie
[tex]C_{n}^{2}=\frac{n!}{(n-2)!*2!}=\frac{(n-1)n}{2}[/tex]
Folosim relatia aceasta pentru prima suma
[tex]S=\frac{1*(1+1)}{2}+\frac{2*(2+1)}{2}+\frac{3*(3+1)}{2}+..+\frac{(n-1)*(n-1+1)}{2}+\frac{n*(n+1)}{2}=\frac{1}{2}(1^{2}+1*1+2^{2}+2*1+3^{2}+3+..+(n-1)^{2}+(n-1)+n^{2}+n)[/tex]
Observi ca se formeaza 2 sume separate al caror rezultat il stim
[tex]S_{1}=1^{2}+2^{2}+3^{2}+..+(n-1)^{2}+n^{2}=\frac{n(n+1)(2n+1)}{6}[/tex]
[tex]S_{2}=1+2+3+..+(n-1)+n=\frac{n(n+1)}{2}[/tex]
Atunci suma devine
[tex]S=\frac{1}{2}(S_{1}+S_{2})=\frac{1}{2}(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2})=\frac{n(n+1)}{2}(\frac{2n+1}{6}+\frac{1}{2})=\frac{n(n+1)}{2}\frac{2n+1+3}{6}=\frac{n(n+1)*2(n+2)}{12}=\frac{n(n+1)(n+2)}{6}[/tex]
b) [tex]S=\frac{2}{1*2}+\frac{2}{2*3}+\frac{2}{3*4}+..+\frac{2}{(n-1)n}+\frac{2}{n(n+1)}2(\frac{1}{1*2}+\frac{1}{2*3}+\frac{1}{3*4}+...+\frac{1}{(n-1)n}+\frac{1}{n(n+1)})[/tex]
Dar observam ca putem scrie
[tex]\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}[/tex]
Atunci putem scrie
[tex]\frac{1}{1*2}=\frac{1}{1}-\frac{1}{2}[/tex]
[tex]\frac{1}{2*3}=\frac{1}{2}-\frac{1}{3}[/tex]
[tex]\frac{1}{3*4}=\frac{1}{3}-\frac{1}{4}[/tex]
----------------------------------------------------------
[tex]\frac{1}{(n-1)*n}=\frac{1}{n-1}-\frac{1}{n}[/tex]
[tex]\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}[/tex]
Facem suma acestor termeni
[tex]\frac{1}{1*2}+\frac{1}{2*3}+\frac{1}{3*4}+..+\frac{1}{(n-1)n}+\frac{1}{n(n+1)}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}+\frac{1}{n}-\frac{1}{n+1}=\frac{1}{1}-\frac{1}{n+1}=\frac{n+1-1}{n+1}=\frac{n}{n+1}[/tex]
