Các câu hỏi tương tự
cmr:1/2!+2/3!+3/4!+...+99/100!<1
20 tháng 9 2023 lúc 15:50
Cho biểu thức : \(C=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}+\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) CMR: \(C< \dfrac{3}{16}\)
CMR:1/2!+2/3!+3/4!+............+99/100!<1
CMR : 1/2! + 2/3! + 3/4! + ... + 99/100! < 1
CMR: \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)
CMR: 100-(\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\))=\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
CMR:
a,\(100\left(1+\frac{1}{2}+\frac{1}{3}+..........+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+........+\frac{99}{100}\)
CMR
1/2! + 2/3! + 3/4! +...+ 99/100! <1
Mn giúp m vs
24 tháng 10 2015 lúc 20:01
CMR: 3^1 + 3^2 +3^3 ...+ 3^99 + 3^100 chia hết cho 4