Các câu hỏi tương tự
Rút gọn:
\(\frac{2016-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{2017}}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{2015}{2016}}\)
Cho \(E=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2016}{^{3^{2016}}}\) CMR \(E< \frac{3}{4}\)
SO SÁNH:
A=\(\frac{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.....+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2016}+\frac{1}{2017}}\)
VÀ
B=2017
Tính nhanh:
A= 1/2+1/2^2+1/2^3+....+1/2^100
B=3^2/2x4+3^2/4x6+3^2/6x8+....+3^2/198x200
C=\(\frac{\frac{2017}{1}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2017}}{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}}\)
D=1x2+2x3+3x4+4x5+...+48x49
E=\(^{1^2+2^2+3^2+...+48^2}\)
F=1x49+2x48+3x47+...+48x2+49x1
A(-1,7).2,3+1,7.(-3,7)-1,7.3-0,17:frac{1}{10}Bleft(frac{1}{2}-1right).left(frac{1}{3}-1right).left(frac{1}{4}-1right)...left(frac{1}{2017}-1right)Cfrac{3}{4}.frac{8}{9}.frac{15}{16}.....frac{899}{900}Dfrac{1}{3}+frac{1}{15}+frac{1}{35}+...+frac{1}{9999}E1-3+3^2-3^3+...+3^{2016}-3^{2017}+3^{2018}G2+2^2+2^3+...+2^{60}
Đọc tiếp
A=(-1,7).2,3+1,7.(-3,7)-1,7.3-0,17:\(\frac{1}{10}\)
B=\(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right).\left(\frac{1}{4}-1\right)...\left(\frac{1}{2017}-1\right)\)
C=\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{899}{900}\)
D=\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9999}\)
E=\(1-3+3^2-3^3+...+3^{2016}-3^{2017}+3^{2018}\)
G=\(2+2^2+2^3+...+2^{60}\)
cho E=\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{2015}{3^{2015}}-\frac{2016}{3^{2016}}\).Chứng minh rằng:E <\(\frac{3}{16}\)
a)Chứng minh rằng: \(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+..+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}=2\)
b)\(A=\frac{-21}{10^{2016}}+\frac{-12}{10^{2017}};B=\frac{-12}{10^{2016}}+\frac{-21}{10^{2017}}\)
So sánh A và B
Cho \(E=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{2015}{3^{2015}}-\frac{2016}{3^{2016}}\) . Chứng minh rằng \(E< \frac{3}{16}\)
Bài cuối đề thi học kỳ 2 môn toán trường mình đó , giải đi mk tk cho.
tính
A=\(\frac{\frac{2017}{2}+\frac{2017}{3}+\frac{2017}{4}+...+\frac{2017}{2018}}{\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}}\)