Question

[nguyen van huy]

Bai 1; a) Cho \(B\) = \(\frac{1}{2}\)+   (\(\frac{1}{2}\))²  +     (\(\frac{1}{2}\))³  +   (\(\frac{1}{2}\))⁴  + ... +  (\(\frac{1}{2}\))⁹⁸  +   (\(\frac{1}{2}\))⁹⁹ . Chứng minh rằng  \(B\) \(< 1\)

          b) Cho \(C\) = \(\frac{1}{3}\)+    \(\frac{1}{3^2}\)+   \(\frac{1}{3^3}\)+...+   \(\frac{1}{3^{98}}\)+   \(\frac{1}{3^{99}}\).  Chứng minh rằng \(C\) \(< \)\(\frac{1}{2}\)

Bai 2;Chứng minh rằng;

           a) \(\frac{3}{1^2.2^2}\)+   \(\frac{5}{2^2.3^2}\)+   \(\frac{7}{3^2.4^2}\)+ ... +   \(\frac{19}{9^2.10^2}\)\(< \)\(1\)

           b) \(\frac{1}{3}\)+  \(\frac{2}{3^2}\)+\(\frac{3}{3^3}\)+ ... +  \(\frac{99}{3^{99}}\)+   \(\frac{100}{3^{100}}\)\(< \frac{3}{4}\)

Answer 1

[Lê Tự Nguyên Hào]

1. a) 2B = 1 + 1/2 + 1/2²+...+1/2⁹⁸

B - B = (1+1/2+...+1/2⁹⁸) - (1/2+....+1/2⁹⁹)

B = 1 - 2⁹⁹ => B < 1

b) Làm tương tự như câu a, ra là (1 - 1/3⁹⁹) : 2 = 1/2 - 1/2.3⁹⁹(C bé hơh 1/2)

Answer 2

[Vũ Quang Vinh]

1. a). Theo đầu bài ta có:
 \(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}\)
\(\Leftrightarrow B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)
\(\Leftrightarrow B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{97}}+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\right)\)
\(\Leftrightarrow B=1-\frac{1}{2^{99}}< 1\)( đpcm )