Question

$\sum_{k=1}^n \cdot 2k = n(n+1) ,n\geq 1$

va rog ajutor

Answer 1

[-] Metoda I - Aducem formula la o expresie cunoscuta a fi adevarata.

Daca expandam suma :

2+4+6+8+....+n=n(n+1)

Daca impartim prin 2 in stanga si in dreapta egalului :

$1+2+3+...+n=\frac{n(n+1)}{2}, \forall n\geq 1$

Am ajuns la formula pentru suma Gauss, formula care stim ca este adevarata.

[-] Metoda II - Inductie matematica

• Verificare

$\text{Daca n = 1 :}\\\\ \sum_{k=1}^{1}(2k)=2=1(1+1)$

• Demonstratie

$\text{Presupunem P(n) :}\\\sum_{k=1}^n2k=n(n+1) \text{ adevarata}, \forall n \in N, n\geq 1\\\\\text{Trebuie sa demonstram P(n+1) : }\\\sum_{k=1}^{n+1}2k = (n+1)(n+2)$

$\sum_{k=1}^{n+1}2k \\\\=(\sum_{k=1}^{n}2k) + 2n \\\\=n(n+1) +2n\\\\=n^2+n+1+2n\\\\=n^2+3n+1\\\\=(n+1)(n+2) \text{, ceea ce trebuia demonstrat.}$