Toán Lớp 6: 1
B=1/31+1/32+1/33+…+1/60 CM rằng 3/5
Toán Lớp 6: 1 B=1/31+1/32+1/33+…+1/60 CM rằng 3/5
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Ta có: 1/31+1/32+1/33+…+1/40>\underbrace{1/40+1/40+1/40+…..+1/40}_{\text{10 phân số}}
⇒1/31+1/32+1/33+…+1/40>10/40=1/4 (1)
Ta có: 1/41+ 1/42+ 1/43+ ….+1/50>\underbrace{1/50+ 1/50+ 1/50+ ….+1/50}_{\text{10 phân số}}
⇒1/41+ 1/42+ 1/43+ ….+1/50>10/50=1/5 (2)
Ta có:1/51+ 1/52+ 1/53+ ….+1/60>\underbrace{1/60+ 1/60+ 1/60+ ….+1/60}_{\text{10 phân số}}
⇒1/51+ 1/52+ 1/53+ ….+1/60>10/60=1/6(3)
Từ (1);(2)(3)
⇒1/31+1/32+1/33+…+1/60>1/4+ 1/5+ 1/6=15/60 + 12/60+ 10/60=37/60>36/60=3/5
Vậy: B>3/5 (**)
Ta có: 1/31+1/32+1/33+…+1/40<\underbrace{1/30+1/30+1/30+…..+1/30}_{\text{10 phân số}}
⇒1/31+1/32+1/33+…+1/40<10/30=1/3 (1)
Ta có: 1/41+ 1/42+ 1/43+ ….+1/50<\underbrace{1/40+ 1/40+ 1/40+ ….+1/40}_{\text{10 phân số}}
⇒1/41+ 1/42+ 1/43+ ….+1/50<10/40=1/4 (2)
Ta có:1/51+ 1/52+ 1/53+ ….+1/60<\underbrace{1/50+ 1/50+ 1/50+ ….+1/50}_{\text{10 phân số}}
⇒1/51+ 1/52+ 1/53+ ….+1/60<10/50=1/5(3)
Từ (1);(2)(3)
⇒1/31+1/32+1/33+…+1/60<1/3+1/4+1/5=20/60+ 15/60 + 12/60=47/60<48/60=4/5
⇒B<4/5 $(**)$
Từ $(*)$ và $(**)$
⇒3/5<B<4/5