\(2A=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{2015}\)
\(\Rightarrow2A-A=1-\left(\frac{1}{2}\right)^{2014}\Rightarrow A=1-\left(\frac{1}{2}\right)^{2014}< 1\)
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{3^2}-1\right).....\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)
\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2013^2}-1\right).\left(\frac{1}{2014^2}-1\right)\)
Cho A= \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\) và B=\(\frac{-1}{2}\). Hãy so sánh A và B
So sánh : A=\(\left(\frac{1}{2^2}\right).\left(\frac{1}{3^2}\right).\left(\frac{1}{4^2}\right)....\left(\frac{1}{2013^2}\right).\left(\frac{1}{2014^2}\right)\)
B = \(-\frac{1}{2}\)
A = \(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2017^2}\right).\)
B = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{6}}{\frac{2015}{1}+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}}\)
Cho \(A=1-\frac{3}{4}+\left(\frac{3}{4}\right)^2-\left(\frac{3}{4}\right)^3+\left(\frac{3}{4}\right)^4-...-\left(\frac{3}{4}\right)^{2013}+\left(\frac{3}{4}\right)^{2014}\)
Chứng minh A không phải là số nguyên
Cho A = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right).....\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\) và B = \(\frac{-1}{2}\). Hãy so sánh A và B
Cho A = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)....\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\) va B = \(\frac{-1}{2}\), So sanh A va B
\(y=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)
x= -1/2
hãy so sánh x và y
