gọi \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{30}}\)
\(\Leftrightarrow2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{29}}\)
\(\Leftrightarrow2A-A=(1+\dfrac{1}{2}+...+\dfrac{1}{2^{29}})-\left(\dfrac{1}{2}+...+\dfrac{1}{2^{30}}\right)\)
\(\Leftrightarrow A=1-\dfrac{1}{2^{30}}< 1\)
đpcm
chứng minh :
a) \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4}\) b) \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2013^2}+\dfrac{1}{2014}>\dfrac{1}{5}\)
chứng minh rằng :
a) \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\) b)\(\dfrac{1}{5^2}+\dfrac{1}{6^5}+...+\dfrac{1}{2013^2}+\dfrac{1}{2014}>\dfrac{1}{5}\)
Chứng minh rằng :
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2013^2}+\dfrac{1}{2014^2}>\dfrac{1}{5}\)
Chứng minh : \(\dfrac{99}{100}\) > \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}>\dfrac{99}{202}\)
chứng minh rằng :
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)
Chứng minh rằng : \(\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+........+\dfrac{1}{2020^2}< \dfrac{1}{4}\)
chứng minh rằng :
b) \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2013^2}+\dfrac{1}{2014}>\dfrac{1}{5}\)
mọi người ơi chú ý hộ mik là cái chỗ \(\dfrac{1}{2014}\) trên kia là đúng nha
Chứng minh rằng :
\(S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{20}}< 1\)
chứng minh ∀ n ϵ N , n > 1 ta có \(\dfrac{1}{n-1}-\dfrac{1}{n}>\dfrac{1}{n^2}>\dfrac{1}{n}-\dfrac{1}{n+1}\)
