Chứng tỏ rằng: 1/3^2+1/4^2+1/5^2+…+1/60^2<4/9 31/08/2021 Bởi aihong Chứng tỏ rằng: 1/3^2+1/4^2+1/5^2+…+1/60^2<4/9
Giải thích các bước giải: Ta có: $\dfrac1{4^2}+\dfrac1{5^2}+…+\dfrac1{60^2}$ $=\dfrac1{4.4}+\dfrac1{5.5}+…+\dfrac1{60.60}$ $<\dfrac1{3.4}+\dfrac1{4.5}+…+\dfrac1{59.60}$ $<\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+…+\dfrac{60-59}{59.60}$ $<\dfrac13-\dfrac14+\dfrac14-\dfrac15+…+\dfrac1{59}-\dfrac1{60}$ $<\dfrac13-\dfrac1{60}$ $<\dfrac13$ $\to \dfrac1{4^2}+\dfrac1{5^2}+…+\dfrac1{60^2}<\dfrac13$ $\to\dfrac1{3^2}+ \dfrac1{4^2}+\dfrac1{5^2}+…+\dfrac1{60^2}<\dfrac1{3^2}+\dfrac13$ $\to\dfrac1{3^2}+ \dfrac1{4^2}+\dfrac1{5^2}+…+\dfrac1{60^2}<\dfrac49$ $\to đpcm$ Bình luận
Giải thích các bước giải:
Ta có:
$\dfrac1{4^2}+\dfrac1{5^2}+…+\dfrac1{60^2}$
$=\dfrac1{4.4}+\dfrac1{5.5}+…+\dfrac1{60.60}$
$<\dfrac1{3.4}+\dfrac1{4.5}+…+\dfrac1{59.60}$
$<\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+…+\dfrac{60-59}{59.60}$
$<\dfrac13-\dfrac14+\dfrac14-\dfrac15+…+\dfrac1{59}-\dfrac1{60}$
$<\dfrac13-\dfrac1{60}$
$<\dfrac13$
$\to \dfrac1{4^2}+\dfrac1{5^2}+…+\dfrac1{60^2}<\dfrac13$
$\to\dfrac1{3^2}+ \dfrac1{4^2}+\dfrac1{5^2}+…+\dfrac1{60^2}<\dfrac1{3^2}+\dfrac13$
$\to\dfrac1{3^2}+ \dfrac1{4^2}+\dfrac1{5^2}+…+\dfrac1{60^2}<\dfrac49$
$\to đpcm$