Chứng tỏ rằng
1/3 +2/32 + 3/33 + 4/34 + … + 100/3100 < 3/4
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Sửa: `4/[3.7]+4/[7.11]+...+4/[95.99]`
`=1/3-1/7+1/7-1/11+....+1/95-1/99`
`=1/3-1/99`
`=32/99`
A = \(\dfrac{4}{3.7}\) + \(\dfrac{4}{7.11}\) + ......+ \(\dfrac{4}{95.99}\)
A = \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\) + ......+ \(\dfrac{1}{95}\) - \(\dfrac{1}{99}\)
A = \(\dfrac{1}{3}\) - \(\dfrac{1}{99}\)
A = \(\dfrac{32}{99}\)
A=\(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}\)
⇒3A=\(1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{99}}\)
⇒\(3A-A=\left(1+\dfrac{2}{3}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}\right)\)
⇒\(2A=1+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\dfrac{100}{3^{100}}\)
⇒\(2A=1+\dfrac{1}{2}-\dfrac{1}{2\cdot3^{99}}-\dfrac{100}{3^{100}}\)
⇒\(A=\dfrac{3}{4}-\dfrac{1}{4\cdot3^{99}}-\dfrac{50}{3^{100}}< \dfrac{3}{4}\)
Vậy......