Question

Calculati suma urmatoare si apoi verificati prin inductie : S= 2+6+10+...+4n+2

Answer 1

[tex]S=2+6+10+...+4n+2\\ S=2+4\cdot 1+2+4\cdot 2+2+...+4n+2\\ S=4(1+2+3+...+n)+2(n+1)\\ S=4\cdot\frac{n(n+1)}{2}+2(n+1)\\ S=2n(n+1)+2(n+1)\\ S=2(n+1)^2\\ Demonstratia\ prin\ inductie:\\ P(n): 2+6+10+...+4n+2=2(0+1)^2\\ i)P(0):2=2(A)\\ ii)Presupunem\ P(k)-A\ \forall k\in N.Se\ demonstreaza\ ca\ si\ P(k+1)\\ este\ adevarat:\\ P(k):2+6+10+..+4k+2=2(k+1)^2\\ P(k+1):\underbrace{2+6+10+...+4k+2}+4(k+1)+2=2(k+2)^2\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~P(k)\\ P(k+1):2(k+1)^2+4k+6=2(k+2)^2\\ P(k+1):2k+3=(k+2)^2-(k+1)^2\\ [/tex]
[tex]P(k+1):2k+3=(k+2-k-1)(k+2+k+1)\\ P(k+1):2k+3=2k+3-A\Rightarrow P(k)-A\Rightarrow P(n)-A. [/tex]