`B=1/31+1/32+1/33+...+1/60`

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`

Đáp án + Giải thích các bước giải:

B = (1/31 + 1/32 + 1/33 + ... + 1/40) + (1/41 + 1/42 + 1/43 + ... + 1/50) + (1/51 + 1/52 + 1/53 + ... + 1/60)

Mà: (1/31 + 1/32 + 1/33 + ... + 1/40) > 1/40 x 10 = 1/4 (Gồm 10 số hạng)

Tương tự: (1/41 + 1/42 + 1/43 + ... + 1/50) > 1/5 ; (1/51 + 1/52 + ... 1/60) > 1/6

B > 1/4 + 1/5 + 1/6

Trong đó: (1/4 + 1/5 + 1/6) > 3/5

Vậy B > 3,5         (1)

Mặt khác:

B = (1/31 + 1/32 + 1/33 + ... + 1/40) + (1/41 + 1/42 + 1/43 + ... + 1/50) + (1/51 + 1/52 + 1/53 + ... + 1/60)

Mà: (1/31 + 1/32 + 1/33 + ... + 1/40) < 1/31 x 10 = 10/30 = 1/3 (gồm 10 số hạng)

Tương tự: (1/41 + 1/42 + 1/43 + ... + 1/50) < 1/4 ; (1/51 + 1/52 + ... + 1/60) < 1/5

Mà B = (1/3 + 1/4 + 1/5) < 4/5 (Vì 1/3 + 1/5 < 3/5 hay 7/12 < 3/5 hay 35/60 < 36/60)

Vậy B < 4,5        (2)

Từ (1) và (2) → 3/5 < B < 4/5