Demonstram egalitatea prin inductie matematica:
[tex]P(n):1+6+20+...+(2n-1)\cdot 2^{n-1}=3+2^{n}\cdot (2n-3)\\
P(1):1=3+2^{1}\cdot (2-3)(Adevarat)\\
Presupunem\ P(n)\ adevarat=>\ demonstram \ P(n+1)\ adevarat\\
P(n+1):1+6+20+...+(2n-1)\cdot 2^{n-1}+(2n+1)\cdot 2^n=\\
=3+2^{n+1}\cdot (2n-1)\\
1+6+20+...+(2n-1)\cdot 2^{n-1}+(2n+1)\cdot 2^n=\\
=3+2^{n}\cdot (2n-3)+(2n+1)\cdot 2^n=3+2^n\cdot (4n-2)=\\
=3+2^{n+1}\cdot (2n-1)
[/tex]
