Đại số lớp 6
Các câu hỏi tương tự
Chứng tỏ rằng:
\(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)
7 tháng 3 2017 lúc 20:34
Chứng minh rằng : \(\dfrac{1}{3}\) - \(\dfrac{2}{3^2}\) + \(\dfrac{3}{3^3}\) - \(\dfrac{4}{3^4}\) + .......... + \(\dfrac{99}{3^{99}}\) - \(\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
Chứng minh rằng \(\dfrac{1}{3}\)-\(\dfrac{2}{3^2}\)+\(\dfrac{3}{3^3}\)-\(\dfrac{4}{3^4}\)+.........+\(\dfrac{99}{3^{99}}\)-\(\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
Cmr : \(\dfrac{1}{3}\) - \(\dfrac{2}{3^2}\) +\(\dfrac{3}{3^3}\) - \(\dfrac{4}{3^4}\) + ...+\(\dfrac{99}{3^{99}}\) - \(\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
CMR: 100- \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}\)
Mọi người giúp em với ạ. em cảm ơn
Chứng tỏ:
\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)=2
CMR 100-(1+\(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\))= (\(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}\))
Tính giá trị biểu thức :
\(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}\right)-\left(\dfrac{1}{5}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{4}{5}\right)+\left(\dfrac{1}{6}+\dfrac{2}{6}+\dfrac{3}{6}+\dfrac{4}{6}+\dfrac{5}{6}\right)-\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{4}{7}+\dfrac{5}{7}+\dfrac{6}{7}\right)+...+\left(100+...+\dfrac{99}{100}\right)\)
CMR \(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}< 1\)