Question

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Answer 1

Răspuns:

Explicație pas cu pas:

Din formula $1+x+x^2+x^3+...+x^n=\frac{x^{n+1}-1}{x-1}$

rezulta ca $2^0+2^1+2^2+...+2^{10}=2^{11}-1$

$\sqrt{2^0+2^1+...+2^{2006}} = \sqrt{2^{2007}}$ dar $2^{2007}$ nu e patrat perfect

deci radicalul de deasupra este irational (adica apartine R - Q)

Demonstratie formula:

$S=1+x+x^2+...+x^n\\xS=x+x^2+x^3+...+x^{n+1}\\xS-S=S(x-1)=x+x^2+x^3+...+x^{n+1}-1-x-x^2-x^3-...-x^n\\S(x-1)=x^{n+1}-1\\S=\frac{x^{n+1}-1}{x-1} \\Deci: 1+x+x^2+...+x^n=\frac{x^{n+1}-1}{x-1}$