Chứng minh rằng:
1) B =\(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}>1\)
2) \(A=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+...+\dfrac{500}{5^{500}}<100\)
3) \(C=\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\dfrac{1}{5^3}+...+\dfrac{1}{500^3}<\dfrac{1}{4}\)
4) \(D=\dfrac{4}{3}+\dfrac{10}{9}+\dfrac{28}{27}+...+\dfrac{3^{98}+1}{3^{98}}<100\)
Làm giúp mình sớm nha! Thanks.
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Đọc tiếp
Chứng minh rằng:
1) B =\(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}>1\)
2) \(A=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+...+\dfrac{500}{5^{500}}<100\)
3) \(C=\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\dfrac{1}{5^3}+...+\dfrac{1}{500^3}<\dfrac{1}{4}\)
4) \(D=\dfrac{4}{3}+\dfrac{10}{9}+\dfrac{28}{27}+...+\dfrac{3^{98}+1}{3^{98}}<100\)
Làm giúp mình sớm nha! Thanks.
lầy dạ??