tính ra nhé bạn. nếu bạn ko bít cách tính
\(2^2.A=\frac{2^2}{2^2}+\frac{1}{2^2}+\frac{1}{2^4}+..+\frac{1}{2^{98}}\)
2^2A-A=3A
\(3A=\frac{2^2}{2^2}-\frac{1}{2^{98}}=1-\frac{1}{2^{98}}<1\)
=> A<1/3=> dpcm
Cho A= \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) và B =\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{200^2}\). Khi đó \(\frac{A}{B}\)= ?
So sánh A ;B : \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2};B=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{200^2}\)
\(A=\left(2+4+6+...+100\right)\left(\frac{3}{5}:0,7+3\left(\frac{-2}{7}\right)\right):\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
Cho \(A=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{100}}\)
Chứng minh rằng \(A< \frac{1}{3}\)
Cho \(A=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{100}}\). Chứng minh \(A< \frac{1}{3}\)
1/ Tính
a) \(P=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
b) Cho \(a+b+c=2010\)và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{3}\)
Tính \(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
2/ Tìm x biết
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}...\frac{30}{62}\cdot\frac{31}{64}=2^x\)
3/ Tìm \(a_1;a_2;a_3;...;a_{100}\)biết \(\frac{a_1-1}{100}=\frac{a_2-2}{99}=\frac{a_3-3}{98}=...=\frac{a_{100}-100}{1}\)và \(a_1+a_2+a_3+...+a_{100}=10100\)
CM\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
So sánh \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2};B=\frac{1}{4^2}+\frac{1}{6^2}+...\frac{1}{200^2^{ }}\)
Cho A=\(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+\frac{1}{2^8}+...+\frac{1}{2^{100}}\). Chứng minh A<\(\frac{1}{3}\)