Question

1.Sa se rezolve urmatoarea inductie matematica:
1²+3²+5²+...+(2n-1)²=n(4n²-1)/3

Answer 1

$\text{Notam propozitia cu P(n)}.\\\text{Etapa 1:Verificarea }\\P(1): 1^2= \dfrac{1\cdot (4\cdot 1^2-1)}{3}\\ ~~~~1= \dfrac{1\cdot 3}{3}\\~~~~~1=1 (A)\\\text{Etapa 2: Demonstratia propriu-zisa }\\\text{Presupunem }P(k)\text{ adevarat }\forall k~ \in~ \mathbb{N^*}.\text{Se demonstreaza ca si P(k+1) }\\\text{este adevarat.}\\P(k): 1^2+3^2+5^2+\ldots +(2k-1)^2= \dfrac{k(4k^2-1)}{3}\\P(k+1): \underbrace{1^2+3^2+5^2+\ldots +(2k-1)^2}+(2k+1)^2= \dfrac{(k+1)(4(k+1)^2-1)}{3}\\~~~~~~~~~~~~~~~~~~~~~~~~~~~P(k)$

$\dfrac{k(4k^2-1)}{3}+(2k+1)^2=\dfrac{(k+1)(4(k+1)^2-1)}{3}  |\cdot 3\\k(4k^2-1)+3(2k+1)^2=(k+1)(2k+2-1)(2k+2+1)\\k(2k-1)(2k+1)+3(2k+1)^2= (k+1)(2k+1)(2k+3)\\(2k+1)[k(2k-1)+3(2k+1)]=(k+1)(2k+1)(2k+3)\\(2k+1)(2k^2-k+6k+3)=(k+1)(2k+1)(2k+3)\\(2k+1)(2k^2-5k+3)=(k+1)(2k+1)(2k+3)\\(2k+1)[2k(k+1)+3(k+1)]=(k+1)(2k+1)(2k+3)\\(k+1)(2k+1)(2k+3)=(k+1)(2k+1)(2k+3) -\text{adevarat}.\\\text{Prin urmare } P(k) \text{ este adevarat },\text{ deci rezulta } P(n)~ \text{adevarat }\forall~ n\in \mathbb{N}^*$