Question

AJUTORRRR Demonstrati prin inducție matematica 1+x+x^2+....+x^n = (x^(n+1) -1 )/ x-1

Answer 1

[tex]1+x+x^2+..+x^n=x^0+x^1+x^2+..+x^n=\frac{x^{n+1}-1}{x-1} [/tex]
$P(0): x^0=\frac{x^{0+1}-1}{x-1}=\frac{x-1}{x-1}=1, (A)$
[tex]P(1):x^0+x^1=\frac{x^{1+1}-1}{x-1}=\frac{x^2-1}{x-1}=\frac{(x+1)(x-1)}{x-1}=x+1, (A) [/tex]
Presupunem 
$P(k):x^0+x^1+...+x^k=\frac{x^{k+1}-1}{x-1}, adev\u arat\u a.$
Demonstram 
[tex]P(k+1):x^0+x^1+x^2+...+x^k+x^{k+1}\\ =(x^0+x^1+x^2+...+x^k)+x^{k+1}=\frac{x^{k+1}-1}{x-1}+x^{k+1}[/tex]
Aducem la acelasi numitor
[tex]=\frac{x^{k+1}-1+x^{k+1}(x-1)}{x-1}=\frac{x^{k+1}-1+x\cdot x^{k+1}-x^{k+1}}{x-1}\\ =\frac{x^{k+1+1}-1}{x-1}=\frac{x^{k+2}}{x-1},[/tex]
c.c.t.d.
Deci
$P(n):1+x+x^2+...+x^n=\frac{x^{n+1}-1}{x-1}$ este adevarata, $\forall n\in \mathbb{N}$