Question

Calculati : P=(1+2)(1+$2 ^{2}$)(1+$2^{4}$)(1+$2^{8}$)(1+$2^{16}$)(1+$2^{32}$)(1+$2^{64}$)=?

Answer 1

Se inmulteste egalitatea cu -1 si obtii:

$-P=(1-2)(1+2)(1+2^2)...(1+2^{64})$

Folosim acum formula

$(a-b)(a+b)=a^2-b^2$  de 7 ori si obtinem

$-P=1-2^{128}\Rightarrow P=2^{128}-1$