Question

1 + 2 + 2^2 + 2^3 + ... + 2^7 =?
cum se calculeaza?

Answer 1

Termenii sunt in progresie geometrica cu b1=2^0 si q=2.
S=b1*(q^n -1)/q-1
S=2^0*(2^8-1)/2-1=1*(2^8-1)/1=2^8-1=256-1=255

Answer 2

   
[tex]\displaystyle\\ \text{Formula generala este: }\\\\ a^0 + a^1 + a^2 +a^3 + ... + a^n = \frac{a^{n+1}-1}{a-1} \\\\ \text{Rezolvare:}\\\\ 1 + 2 + 2^2 + 2^3 + ... + 2^7 = \frac{2^{7+1}-1}{2-1} = \frac{2^{8}-1}{1} = \boxed{\bf 2^{8}-1}[/tex]