Question

Exericitiul 7, subpunctul b). cum il rezolv prin inductie ? et de dem nu imi iese

Answer 1

se verifica pt n=2
2^3-1/2^3+1=7/9=2*(2^2+2+1)/3*2*(2+1)=2*7/18=7/9
pt n=2 e avevarat
pres ca pt n=k e adevarat
2^3-1/2^3+1*......*k^3-1/K^3+1=2*(k^2+k+1)/3*k(k+1)
trebuie sa dem ca e adevarat si pt k+1
2*(k^2+k+1)/3*k*(k+1)*[(k+1)^3-1][(k+1)^3+1]
se folosesc formulele a^3+b^3=(a+b)(a^2-ab+b^2)
a^3-b^3=(a-b)(a^2+ab+b^2)
se adapteaza pt [(k+1)^3-1] si pentru [(k+1)^3+1]
$\frac{2(k^2+k+1)}{3k(k+1)} * \frac{[(k+1)-1][(k+1)^2+(k+1)+1]}{[(k+1)+1][(k+1)^2-(k+1)+1]}} = \frac{2(k^2+k+1)}{3k(k+1)} * \frac{k(k^2+3k+3)}{(k+2)(k^2+k+1)} =$
se simplifica k^2+k+1 si k si devine

$\frac{2(k^2+3k+3)}{3(k+1)(k+2)}$     (1) 
 in expresia  $\frac{2(n^2+n+1)}{3n(n+1)}$ se inlocuieste n cu k+1 si vedem ca expresia devine

$\frac{2[(k+1)^2+(k+1)+1]}{3(k+1)(k+2)} = \frac{2(k^2+3k+3)}{3(k+1)(k+2)}$)


exact ceea  ce aveam in expresia (1)