Các câu hỏi tương tự
a,tính:A=\(\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right).....\left(\frac{1}{100}-1\right).\left(\frac{1}{121}-1\right)\)
b,tình=\(2^{2010}-2^{2009}-2^{2008}-...-2-1\)
giúp với!!!!!♫♪
cho A= 1/2 + 1/3 + 1/4 +...+ 1/2020 và B= 2019/1 + 2018/2 + 2017/3 +...+ 1/2019. Tính A/B
Tính tỉ số A/B biết:
A=1/2 + 1/3 + 1/4 + ... + 1/2017 + 1/2018 + 1/2019
B=2018/1 + 2017/2 + 2016/3 + ... + 2/2017 + 1/2018
Tính.
a)A=1/1.3.5+1/3.5.7+1/5.7.9+...+1/97.99.101
b)B=2^2019-2^2018-2^2017-...-2+1
c)C=1/1.2+1/3.4+...+1/49.50-1/50-1/49-...-1_2
d)D=1/2!+5/3!+11/4!+..+99.100-1/100!
Tính nhanh:
A=(-1/2)-(-3/5)+(-1/9)+1/131-(-2/7)+4/35-7/18
B=(1/4-1).(1/9-1).(1/16-1)....(1/121-1)
C=1/1.3-1/3.5-.....-1/49.51
A=1-1/2+1/3-1/4+1/5-...-1/2018+1/2019
và B=1/1010+1/1011+...+1/2019
Tính ( A-B-1)^2019
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)]
Tínha)Aleft(1-frac{1}{1+2}right)left(1-frac{1}{1+2+3}right)...left(1-frac{1}{1+2+3+...+2006}right)b)B1+frac{3}{2^3}+frac{4}{2^4}+...+frac{100}{2^{100}}c)Cfrac{1}{2!}+frac{2}{3!}+...+frac{n-1}{n!}d)D1+2^2+3^2+...+98^2e)E3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1f)F2^{2010}-2^{2009}-...-2-1g)Gleft(frac{1}{4}-1right)left(frac{1}{9}-1right)left(frac{1}{16}-1right)...left(frac{1}{100}-1right)left(frac{1}{121}-1right)
Đọc tiếp
Tính
a)A=\(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2006}\right)\)
b)\(B=1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\)
c)C=\(\frac{1}{2!}+\frac{2}{3!}+...+\frac{n-1}{n!}\)
d)D=\(1+2^2+3^2+...+98^2\)
e)E=\(3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\)
f)F=\(2^{2010}-2^{2009}-...-2-1\)
g)G=\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{100}-1\right)\left(\frac{1}{121}-1\right)\)
A=1-\(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}+\dfrac{1}{2019}\)
B=\(\dfrac{1}{1010}+\dfrac{1}{1011}+\dfrac{1}{1012}+...+\dfrac{1}{2019}\)
Tính \(^{\left(A-B\right)^{2019}}\)