Question

05. Să se demonstreze prin inductie matematică egalitatea 1x5+2x8+...+n(3n+2)=(n(n+1)(2n+3))/2​

Answer 1

$\displaystyle\it\\\underline{etapa~1~:~verificare}.\\\\P(1):~3\cdot1+2=\frac{n(n+1)(2n+3)}{2} =5~~(A).\\\\\underline{etapa~2~:P(k)\implies P(k+1)}.\\\\P(k)=5+16+..+k(3k+2)~~(A).\\P(k+1)=\underbrace{5+16+...+k(3k+2)}_{P(k)}+(k+1)\Big[3(k+1)+2\Big] =\\\frac{k(k+1)(2k+3)}{2} +3k^2+8k+5=\frac{k(k+1)(2k+3)+6k^2+16k+10}{2} =\\\frac{(\underline{k+1})(\underline{k+1}+1)(2(\underline{k+1})+3)}{2},~(qed).$