Question

Chứng minh rằng:

\(\dfrac{7}{1^3.2^3}+\dfrac{19}{2^3.3^3}+\dfrac{37}{3^3.4^3}+...+\dfrac{29701}{99^3.100^3}< 1\)

Answer 1

\(\dfrac{7}{1^3\cdot2^3}+\dfrac{19}{2^3\cdot3^3}+\dfrac{37}{3^3\cdot4^3}+...+\dfrac{29701}{99^3\cdot100^3}\\ =\dfrac{2^3-1^3}{1^3\cdot2^3}+\dfrac{3^3-2^3}{2^3\cdot3^3}+\dfrac{4^3-3^3}{3^3\cdot4^3}+...+\dfrac{100^3-99^3}{99^3\cdot100^3}\\ =\dfrac{2^3}{1^3\cdot2^3}-\dfrac{1^3}{1^3\cdot2^3}+\dfrac{3^3}{2^3\cdot3^3}-\dfrac{2^3}{2^3\cdot3^3}+...+\dfrac{100^3}{99^3\cdot100^3}-\dfrac{99^3}{99^3\cdot100^3}\\ =\dfrac{1}{1^3}-\dfrac{1}{2^3}+\dfrac{1}{2^3}-\dfrac{1}{3^3}+...+\dfrac{1}{99^3}-\dfrac{1}{100^3}\\ =1-\dfrac{1}{100^3}< 1\)

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