Fie :
[tex]P(k):(1- \frac{1}{2^{2}} )(1- \frac{1}{3^{2}} )...(1- \frac{1}{(k+1)^{2}} )= \frac{k+2}{2k+2} [/tex]
Verificam un caz particular:
P(1): (1 - 1/2²) = (1 + 2)(2 + 2) ==> 1 - 1/4 = 3/4 (Adevarat)
P(k) implica P(k + 1), deci daca P(k) este adevarata, atunci si P(k + 1) trebuie sa fie adevarata
Presupunem P(k) adevarata:
[tex]P(k):(1- \frac{1}{2^{2}} )(1- \frac{1}{3^{2}} )...(1- \frac{1}{(k+1)^{2}} )= \frac{k+2}{2k+2} [/tex]
[tex]P(k+1):(1- \frac{1}{2^{2}} )(1- \frac{1}{3^{2}} )...(1- \frac{1}{(k+1)^{2}} )(1- \frac{1}{(k+2)^2} )= \frac{k+3}{2k+4}[/tex]
Dar daca ne uitam la P(k + 1), observam ca toti termenii imultirii, mai putin ultimul sunt defapt P(k), si putem inlocui cu ce stim deja:
[tex]P(k+1): \frac{k+2}{2k+2}(1- \frac{1}{(k+2)^2} )= \frac{k+3}{2k+4} \\
\frac{k+2}{2k+2}- \frac{1}{(2k+2)(k+2)}= \frac{k+3}{2k+4}\\
\frac{(k+2)^2}{(2k+2)(k+2)} - \frac{1}{(2k+2)(k+2)}= \frac{k+3}{2k+4}\\
\frac{k^2+4k+4-1}{2(k+1)(k+2)}= \frac{k+3}{2(k+2)}\\
\frac{k^2+k+3k+3}{2(k+1)(k+2)} = \frac{k+3}{2(k+2)}\\
\frac{k(k+1)+3(k+1)}{2(k+1)(k+2)} = \frac{k+3}{2(k+2)}\\
\frac{(k+1)(k+3)}{2(k+1)(k+2)} = \frac{k+3}{2(k+2)}\\
[/tex]
Care este adevarata ==> Presupunerea facuta este adevarata ==> Propozitia este mereu adevarata
