a) Ta có
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^6}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)\)
\(A=1-\frac{1}{2^7}\)
Do \(1-\frac{1}{2^7}< 1\Rightarrow A< 1\left(đpcm\right)\)
Chứng minh rằng :\(\frac{1}{3^1}+\frac{2}{3^2}+\frac{3}{3^3}+.....+\frac{100}{3^{100}}+\frac{101}{3^{101}}< \frac{3}{4}\)
Nhanh lên nhé . Mk đang cần gấp
Tính các tổng sau:
a) \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}.\)
b) \(-\frac{4}{5}+\frac{4}{5^2}-\frac{4}{5^3}+...+\frac{4}{5^{200}}.\)
c)\(\frac{-1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)
CMR: \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{101}{3^{101}}+\frac{102}{3^{102}}<\frac{3}{4}\)
CMR:
a, \(100-\left(1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+..+\frac{99}{100}\)
b, \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+..+\frac{1}{200}\right)=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Giải nhanh giùm mình nhé!!!!!!!!!!!!!!
Tính : \(K=\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right).3^5+\left(\frac{1}{3^5}+\frac{1}{3^6}+\frac{1}{3^7}+\frac{1}{3^8}.3^9\right)+...+\left(\frac{1}{3^{97}}+\frac{1}{3^{98}}+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right).3^{101}\)
Cho biểu thức D =\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^2}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\) chứng minh rằng D < \(\frac{3}{4}\)
A =\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{101}{3^{101}}\)
Chứng minh A < \(\frac{3}{4}\)
\(\frac{1}{2}-1+\frac{3}{2}-\frac{4}{2}+\frac{5}{2}-...-\frac{100}{2}+\frac{101}{2}\)=?
A=\(\frac{-3}{5}+\left(\frac{-2}{5}+2\right)\)
\(B=\left(6-2\frac{4}{5}\right)\times3\frac{1}{8}-1\frac{3}{5}=\frac{1}{4}\)
Tính nhanh
\(A=\frac{5}{1\times7}+\frac{5}{4\times7}+\frac{5}{7\times10}+......+\frac{5}{101\times104}\)
