Câu hỏi của Nguyen Thu Ngoc

Question

Tinh :P= 1/2 +1/2^2 + ... + 1/ 2^99

ai lam duoc minh tich cho

Thắng Nguyễn · Accepted Answer

$2P=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)$

$2P=1+\frac{1}{2}+...+\frac{1}{2^{98}}$

$2P-P=\left(1+\frac{1}{2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)$

$P=1-\frac{1}{2^{99}}$

Câu trả lời:

$2P=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)$

$2P=1+\frac{1}{2}+...+\frac{1}{2^{98}}$

$2P-P=\left(1+\frac{1}{2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)$

$P=1-\frac{1}{2^{99}}$

VICTORY_ Trần Thạch Thảo · Answer

$P=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}$

$2P=2.\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)$

$2P=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99-1}}$

$2P=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}$

$P=2P-P=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)$

$P=1-\frac{1}{2^{99}}$

$P=\frac{2^{99}-1}{2^{99}}$

Câu trả lời:

$P=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}$

$2P=2.\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)$

$2P=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99-1}}$

$2P=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}$

$P=2P-P=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)$

$P=1-\frac{1}{2^{99}}$

$P=\frac{2^{99}-1}{2^{99}}$