\(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}=1-\frac{1}{2016}=\frac{2015}{2016}< \frac{1512}{2016}=\frac{3}{4}\)
Vậy \(A< \frac{3}{4}\)
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
1/Tính nhanh
P=(1-\(\frac{1}{2^2}\)) x (1-\(\frac{1}{3^2}\)) x (1-\(\frac{1}{4^2}\)) x ... x (1-\(\frac{1}{50^2}\))
2/Cho Q=(1-\(\frac{1}{2^2}\)) x (1-\(\frac{1}{3^2}\)) x (1-\(\frac{1}{4^2}\)) x ... x (1-\(\frac{1}{40^2}\)) . So sánh Q với \(\frac{1}{2}\)
3/So sánh: A = \(\frac{2016^{2016}+1}{2016^{2017}+1}\)và B = \(\frac{2016^{2017}-3}{2016^{2018}-3}\)
so sánh
A = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2016}}với\frac{3}{2}\)
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 tổng T = \(\frac{2}{2^1}\)+ \(\frac{3}{2^2}\)+ \(\frac{4}{2^3}\)+ ... + \(\frac{2016}{2^{2015}}\)+ \(\frac{2017}{2^{2016}}\)
So sánh T với 3
Cho tổng T=\(\frac{2}{2^1}\)+\(\frac{3}{2^2}\)+\(\frac{4}{2^3}\)+...+\(\frac{2016}{2^{2015}}\)+\(\frac{2017}{2^{2016}}\)
So sánh T với 3.
so sánh: S=\(\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+...+\frac{2016}{2017!}\)
choa=\(\frac{2}{2^1}+\frac{3}{2^2}+...+\frac{2017}{2^{2016}}\)
so sánh a với 3
T=\(\frac{2}{2^1}\)+ \(\frac{3}{2^2}\)+\(\frac{4}{2^3}\)+...+\(\frac{2017}{2^{2016}}\). So sánh T với 3
