CMR:\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^{2016}}< 2016\)
Những câu hỏi liên quan
CMR:\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^{2016}}< 2016\)
\(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^{2006}}\)
\(\Rightarrow A< 1+1+1+...+1\)
\(\Rightarrow A< 2016\)
Cho \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017};B=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}\).CMR B/A là số nguyên
Ta có :
\(B=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}\)
\(B=\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{1}{2016}+1\right)+1\)
\(B=\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2016}+\frac{2017}{2017}\)
\(B=2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)\)
\(\Rightarrow\frac{B}{A}=\frac{2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}=2017\)
Vậy \(\frac{B}{A}\)là số nguyên
Đúng 0
Bình luận (0)
cmr:
S= \(\frac{1}{5^1}+\frac{2}{5^2}+...+\frac{2016}{5^{2016}}< \frac{1}{3}\)
Với B =1-\(\frac{1}{2^2}\)-\(\frac{1}{3^2}\)-....................-\(\frac{1}{2016^2}\)>\(\frac{1}{2016}\)CMR B >\(\frac{1}{2016}\)
ngai viet qua , biet lam nhung ko viet dau
Đúng 0
Bình luận (0)
CMR: \(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^2}+...+\frac{2016}{2^{2016}}<2\)
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2^{2016}-2}+\frac{1}{2^{2016}-1}>1008\)
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2^{2016}-2}+\frac{1}{2^{2016}-1}>1008\)
tinh B=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2016}}{\frac{2016}{1}+\frac{2003}{2}+\frac{2002}{3}+...+\frac{1}{2016}}\)
Tính nhanh : \(\frac{2017+\frac{1}{2016}+\frac{2}{2015}+\frac{3}{2014}+...+\frac{2015}{2}+\frac{2016}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}}\)
CMR:
\(\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{2016^3}< \frac{1}{4}\)