Question

[peter canner]

\(\frac{1}{2}\)+\(\frac{1}{6}\)+\(\frac{1}{12}\)+\(\frac{1}{20}\)+\(\frac{1}{30}\)+\(\frac{1}{42}\)+\(\frac{1}{56}\)+\(\frac{1}{72}\)+\(\frac{1}{90}\)+\(\frac{1}{110}\)=

\(\frac{1}{2}\)+\(\frac{1}{4}\)+\(\frac{1}{8}\)+\(\frac{1}{16}\)+\(\frac{1}{32}\)+\(\frac{1}{64}\)+\(\frac{1}{128}\)=

3 tik 1 cau 3 nguoi nhanh nhat

Answer 1

[Xyz OLM]

a)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\)

= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{10.11}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)

= \(1-\frac{1}{11}\)

= \(\frac{10}{11}\)

b) Đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)

= \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\)

=> 2A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\)

Lấy 2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)

              A  = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^7}\)

              A  = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^2}-...-\frac{1}{2^6}+\frac{1}{2^6}-\frac{1}{2^7}\)

             A   =\(1-\frac{1}{2^7}\)

Answer 2

[Nguyễn Tấn Phát]

Đặt \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}+\frac{1}{110}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}+\frac{1}{10.11}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

\(A=1-\frac{1}{11}\)

\(A=\frac{10}{11}\)

Đặt \(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)

\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\left(1\right)\)

\(2B=\frac{2}{2}+\frac{2}{2^2}+\frac{2}{2^3}+\frac{2}{2^4}+\frac{2}{2^5}+\frac{2}{2^6}+\frac{2}{2^7}\)

\(2B=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\left(2\right)\)

Lấy \(\left(2\right)-\left(1\right)\)hay \(2B-B\)ta có:

\(2B-B=\left(1+\frac{1}{2}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)\)

\(\Rightarrow B=1-\frac{1}{2^7}\)

\(\Rightarrow B=\frac{2^7-1}{2^7}=\frac{128-1}{128}=\frac{127}{128}\)

HOK TOT