Các câu hỏi tương tự
Chứng tỏ : \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2019}{3^{2019}}< 0,75\)
chứng tỏ: 1/3+2/3^2+......+2019/362019<0,75
cho a = 1 - 2019 /2020 + ( 2019/2020)^2 -(2019-2020)^3 +....+(2019/2020) ^2020 chứng tỏ a ko phải là một số nguyên
1chứng tỏ 1/3+2/3^2 +3/3^3+...+2019/3^2019 < 0,75
2 . S=2^100 - 2^99 + 2^98+....+2^2-2
mk cần gấp nha
Tính :A= [(2018/1)+(2017/2)+(2016/3)+(2015/4)+...+(4/2015)+(3/2016)+(2/2017)+(1/2018)]/[(2019/1)+(2019/2)+(2019/3)+(2019/4)+...+(2019/2015)+(2019/2016)+(2019/2017)+(2019/2018)+(2019/2019)]
Tính :A= [(2018/1)+(2017/2)+(2016/3)+(2015/4)+...+(4/2015)+(3/2016)+(2/2017)+(1/2018)]/[(2019/2)+(2019/3)+(2019/4)+(2019/5)+...+(2019/2015)+(2019/2016)+(2019/2017)+(2019/2018)+(2019/2019)]
chứng tỏ rằng
a) A= \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}< 1\)
b) B= \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2019}}< \frac{1}{2}\)
Chứng tỏ rằng x-1/2^2-1/3^2-1/4^2-...-1/2019^2-1/2020^2=0,x không là số nguyên
cho A = 1 -2/3 +(2/3)^2-(2/3)^3+.....+ (2/3)^2018-(2/3)^2019 chứng tở A không nguyên