Question

Tính giá trị của P, biết rằng:

P = 1/256 + 1/128 + 1/64 + ... + 1

Answer 1

\(P=\dfrac{1}{256}+\dfrac{1}{128}+...+1\)

\(=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^8\)

=>\(2P=2+1+\dfrac{1}{2}+...+\left(\dfrac{1}{2}\right)^7\)

=>\(2P-P=2+1+\dfrac{1}{2}+...+\left(\dfrac{1}{2}\right)^7-1-\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2-...-\left(\dfrac{1}{2}\right)^8\)

=>\(P=2-\left(\dfrac{1}{2}\right)^8=\dfrac{511}{256}\)