Toán chứng minh 1/3^2+1/4^2+1/5^2+…+1/60^2<4/9 13/07/2021 By Vivian chứng minh 1/3^2+1/4^2+1/5^2+…+1/60^2<4/9
$\begin{array}{l} \dfrac{1}{{{3^2}}} + \dfrac{1}{{{4^2}}} + \dfrac{1}{{{5^2}}} + … + \dfrac{1}{{{{60}^2}}}\\ < \dfrac{1}{9} + \dfrac{1}{{3.4}} + \dfrac{1}{{4.5}} + \dfrac{1}{{5.6}} + … + \dfrac{1}{{59.60}}\\ = \dfrac{1}{9} + \dfrac{1}{3} – \dfrac{1}{4} + \dfrac{1}{4} – \dfrac{1}{5} + … + \dfrac{1}{{59}} – \dfrac{1}{{60}}\\ = \dfrac{1}{9} + \dfrac{1}{3} – \dfrac{1}{{60}} = \dfrac{4}{9} – \dfrac{1}{{60}} < \dfrac{4}{9} \end{array}$ Trả lời
`1/3^2+1/4^2+1/5^2+…+1/60^2<4/9` $=\dfrac1{4.4}+\dfrac1{5.5}+…+\dfrac1{60.60}$ $<\dfrac1{3.4}+\dfrac1{4.5}+…+\dfrac1{59.60}$ $<\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+…+\dfrac{60-59}{59.60}$ $<\dfrac13-\dfrac14+\dfrac14-\dfrac15+…+\dfrac1{59}-\dfrac1{60}$ $<\dfrac13-\dfrac1{60}$ $<\dfrac13$ $\to \dfrac1{4^2}+\dfrac1{5^2}+…+\dfrac1{60^2}<\dfrac13$ $\to\dfrac1{3^2}+ \dfrac1{4^2}+\dfrac1{5^2}+…+\dfrac1{60^2}<\dfrac1{3^2}+\dfrac13$ $\to\dfrac1{3^2}+ \dfrac1{4^2}+\dfrac1{5^2}+…+\dfrac1{60^2}<\dfrac49$ Vậy `…. < 4/9` Trả lời
$\begin{array}{l} \dfrac{1}{{{3^2}}} + \dfrac{1}{{{4^2}}} + \dfrac{1}{{{5^2}}} + … + \dfrac{1}{{{{60}^2}}}\\ < \dfrac{1}{9} + \dfrac{1}{{3.4}} + \dfrac{1}{{4.5}} + \dfrac{1}{{5.6}} + … + \dfrac{1}{{59.60}}\\ = \dfrac{1}{9} + \dfrac{1}{3} – \dfrac{1}{4} + \dfrac{1}{4} – \dfrac{1}{5} + … + \dfrac{1}{{59}} – \dfrac{1}{{60}}\\ = \dfrac{1}{9} + \dfrac{1}{3} – \dfrac{1}{{60}} = \dfrac{4}{9} – \dfrac{1}{{60}} < \dfrac{4}{9} \end{array}$
`1/3^2+1/4^2+1/5^2+…+1/60^2<4/9`
$=\dfrac1{4.4}+\dfrac1{5.5}+…+\dfrac1{60.60}$
$<\dfrac1{3.4}+\dfrac1{4.5}+…+\dfrac1{59.60}$
$<\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+…+\dfrac{60-59}{59.60}$
$<\dfrac13-\dfrac14+\dfrac14-\dfrac15+…+\dfrac1{59}-\dfrac1{60}$
$<\dfrac13-\dfrac1{60}$
$<\dfrac13$
$\to \dfrac1{4^2}+\dfrac1{5^2}+…+\dfrac1{60^2}<\dfrac13$
$\to\dfrac1{3^2}+ \dfrac1{4^2}+\dfrac1{5^2}+…+\dfrac1{60^2}<\dfrac1{3^2}+\dfrac13$
$\to\dfrac1{3^2}+ \dfrac1{4^2}+\dfrac1{5^2}+…+\dfrac1{60^2}<\dfrac49$
Vậy `…. < 4/9`