Răspuns :
[tex]\text{Demonstram prin inductie:}\\\text{Pasul 1: Verificarea }\\P(1):\left(1-\dfrac{1}{2^2}\right)=\dfrac{1+2}{2(1+1)}\\1-\dfrac{1}{4}=\dfrac{3}{2\cdot 2}\\\dfrac{4-1}{4}=\dfrac{3}{4}\\\dfrac{3}{4}=\dfrac{3}{4}-\bold{adevarat}[/tex]
[tex]\text{Pasul 2:Demonstratia propriu-zisa:}\\\text{Presupunem P(k)},\forall k\in \mathbb{N}^*.\text{Demonstram ca si }P(k+1)\text{ este adevarat.}\\P(k): \left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)\ldots \left(1-\dfrac{1}{(n+1)^2}\right)=\dfrac{k+2}{2(k+1)}\\P(k+1): \underbrace{\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\ldots \left(1-\dfrac{1}{(k+1)^2}\right)}\left(1-\dfrac{1}{(n+2)^2}\right)=\dfrac{k+3}{2(k+2)}\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~P(k)[/tex]
[tex]\dfrac{k+2}{2(k+1)}\left(1-\dfrac{1}{(k+2)^2}\right)=\dfrac{k+3}{2(k+2)}\\\dfrac{k+2}{2(k+1)}\cdot \dfrac{(k+2)^2-1}{(k+2)^2}=\dfrac{k+3}{2(k+2)}\\\dfrac{(k+2-1)(k+2+1)}{2(k+1)(k+2)}=\dfrac{k+3}{2(k+2)}\\\dfrac{(k+1)(k+3)}{2(k+1)(k+2)}=\dfrac{k+3}{2(k+2)}\\\dfrac{k+3}{2(k+2)}=\dfrac{k+3}{2(k+2)}-\bold{adevarat}\\\text{Prin urmare P(k) este adevarat,de unde si P(n) este adevarat}~ \forall~ n \in \mathbb{N}^*[/tex]
