P(1) : n=1 1²=[tex] \frac{1 ori 2 ori 3}{6} [/tex]
1=1 (A)
P(2) : n=2 1²+2²=[tex] \frac{2 ori 3 ori 5}{6} [/tex]
5=5 (A)
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P(k) : n=k 1²+2²+3²+...+k² = [tex] \frac{k(k+1)(2k+1)}{6} [/tex]
P(k+1) : n=k+1
1²+2²+3²+...+k² + (k+1)² = [tex] \frac{(k+1)(k+2)(2(k+1)+1)}{6} [/tex]
[tex] \frac{k(k+1)(2k+1)}{6} [/tex] + (k+1)² = [tex] \frac{(k+1)(k+2)(2k+3)}{6} [/tex]
[Amplificam (k+1)² cu 6 pentru a avea numitor comun]
[tex] \frac{k(k+1)(2k+1)+ 6(k+1)²}{6} [/tex] =[tex] \frac{(k+1)(k+2)(2k+3)}{6} [/tex]
[In membrul stang dam factor comun pe (k+1)]
[tex] \frac{(k+1)(k(2k+1)+ 6(k+1)}{6} [/tex] = [/tex] =[tex] \frac{(k+1)(k+2)(2k+3)}{6} [/tex]
[tex] \frac{(k+1)(2k² + k + 6k +6)}{6} [/tex] = [/tex] =[tex] \frac{(k+1)(2k² + 3k + 4k +6)}{6} [/tex]
[tex] \frac{(k+1)(2k² + 7k + 6)}{6} [/tex] = [tex] \frac{(k+1)(2k² + 7k + 6)}{6} [/tex] (A)
* unde acel A la a 2-a inseamna numar la patrat
ex: 2k A2 = 2k la patrat
