Question

Sa se demonstreze prin metoda inductiei matematice complete ca :

$1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$ , ∀ n ∈ $N^*$

Answer 1

Răspuns:

Explicație pas cu pas:

$1.~pentru~n=1,~1^{2}=\frac{1*(1+1)*(2*1+1)}{6}=\frac{1*2*3}{6} =1,~Adevarat\\2.~Admitem~ca~pt.~n=k,~1^{2}+2^{2}+...+k^{2}=\frac{k*(k+1)*(2k+1)}{6}~e~Adevarat\\3.~Verificam~daca~afirmatia~(egalitatea)~e~adevarata~pentru~n=k+1\\1^{2}+2^{2}+...+k^{2}+(k+1)^{2}=\frac{(k+1)*(k+1+1)*(2(k+1)+1)}{6}=\frac{(k+1)*(k+2)*(2k+3)}{6}\\1^{2}+2^{2}+...+k^{2}+(k+1)^{2}=\frac{k*(k+1)*(2k+1)}{6}+(k+1)^{2}=(k+1)*(\frac{k*(2k+1)}{6}+k+1)=(k+1)*\frac{2k^{2}+k+6k+6}{6} =(k+1)*\frac{2k^{2}+7k+6}{6}=$

=(k+1)*2*(k-(-2))(k-(-3/2))/6=(k+1)(k+2)(2k+3)/6 am ajuns la ce trebuia.. :))

Deci egalitatea e afevarata pentru orice n, natural, n>0.

am descompus trinomul 2k²+7k+6 in factori dupa formula:

ax²+bc+c=a(x-x1)(x-x2), folosind delta