Lời giải:
Ta có:
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+....+\left(\frac{1}{2}\right)^{99}\)
\(\Rightarrow \frac{1}{2}B=\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+....+\left(\frac{1}{2}\right)^{100}\)
Trừ theo vế:
\(\Rightarrow \frac{B}{2}=\left(\frac{1}{2}\right)^{100}-\frac{1}{2}\)
\(\Leftrightarrow B=\left(\frac{1}{2}\right)^{99}-1<2-1\Leftrightarrow B< 1\)
Vì \(\left(\frac{1}{2}\right)^{99}\not\in\mathbb{Z};1\in\mathbb{Z}\Rightarrow B\not\in \mathbb{Z}\)
Ta có đpcm.
\(PHUCDZ=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}\)
\(PHUCDZ=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)
\(2PHUCDZ=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)\)
\(2PHUCDZ=1+\dfrac{1}{2}+...+\dfrac{1}{2^{98}}\)
\(2PHUCDZ-PHUCDZ=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^{98}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)\)
\(PHUCDZ=1-\dfrac{1}{2^{99}}< 1\)
\(\Rightarrowđpcm\)
\(PHUCDZ=1-\dfrac{1}{2^{99}}=\dfrac{2^{99}}{2^{99}}-\dfrac{1}{2^{99}}=\dfrac{2^{99}-1}{2^{99}}\)
Vì \(2^{99}-1\) và \(2^{99}\) là 2 số nguyên tố cùng nhau nên không thể rút gọn cho 1 số nào khác 1.
Vậy \(PHUCDZ\ne Z\Rightarrowđpcm\)
