a: \(2B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)
=>\(2B-B=1-\dfrac{1}{2^{100}}=\dfrac{2^{100}-1}{2^{100}}\)
=>\(B=\dfrac{2^{100}-1}{2^{100}}\)
b: \(2C=1-\dfrac{1}{2}+...+\dfrac{1}{2^{98}}-\dfrac{1}{2^{99}}\)
=>\(3C=1-\dfrac{1}{2^{100}}=\dfrac{2^{100}-1}{2^{100}}\)
=>\(C=\dfrac{2^{100}-1}{2^{100}\cdot3}\)
tính D =\(\dfrac{100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+....+\dfrac{99}{100}}\)
Chứng minh \(\dfrac{1}{5}\)< \(\dfrac{1}{4^2}\)+\(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)+\(\dfrac{1}{7^2}\)+......+\(\dfrac{1}{99^2}\)+\(\dfrac{1}{100^2}\)<\(\dfrac{1}{3}\)
chứng tỏ rằng
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}>\dfrac{99}{202}\)
Chứng minh rằng :
\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}=2\)
Chứng tỏ rằng :\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)=2
chứng tỏ rằng
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
CMR:A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.........+\dfrac{1}{100^2}< 1\)
Cho \(B=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}\). Chứng minh \(B< 1\dfrac{3}{4}\)
Chứng tỏ rằng:\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}=2\)
Trình bày ra dùm mình nha!!Giúp em nha!!!
