Ne uitam la numarator si observam regula
[tex]3=2*1+1[/tex]
[tex]5=2*2+1[/tex]
[tex]7=2*3+1[/tex]
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[tex]2*n+1=2*n+1[/tex]
Adunam toate aceste egalitati
[tex]3+5+7+...+(2n+1)=2*1+1+2*2+1+2*3+1+..+2*n+1=2(1+2+3+..+n)+1*n[/tex]
Ca sa obtinem numaratorul mai adunam in stanga si in dreapta cu 1
[tex]1+3+5+7+...+(2*n+1)=1+n+2(1+2+3+..+n)[/tex]
Mai stim ca
[tex]1+2+3+..+n=\frac{n(n+1)}{2}[/tex]
Atunci(notand numaratorul cu N)
[tex]N=n+1+2*\frac{n(n+1)}{2}=(n+1)+n(n+1)=(n+1)(1+n)=(n+1)^{2}[/tex]
Sa ne ultam si numitorul ce formulare are
[tex]2(1+2+3+..+n+1)=2*\frac{(n+1)(n+2)}{2}=(n+1)(n+2)[/tex]
Atunci fractia devine
[tex]\frac{(n+1)^{2}}{(n+1)(n+2)}=\frac{n+1}{n+2}[/tex] Deci fractia este reductibila
e) [tex]\frac{n^{2}+3n+2}{n^{2}+4n+3}=\frac{n^{2}+n+2n+2}{n^{2}+n+3n+3}=\frac{n(n+1)+2(n+1)}{n(n+1)+3(n+1)}=\frac{(n+1)(n+2)}{(n+1)(n+3)}=\frac{n+2}{n+3}[/tex] deci evident fractia este reductibila.
