Đáp án:

 $\frac{1}{4}$ < S < $\frac{91}{330}$  (dpcm)

Giải thích các bước giải:

 +) S= $\frac{1}{101}+$ $\frac{1}{102}$ +...+ $\frac{1}{130}$

   =( $\frac{1}{101}$ + $\frac{1}{102}$ + ... + $\frac{1}{110)}$) + ($\frac{1}{111}$ + $\frac{1}{112}$ + ... + $\frac{1}{120}$ ) + ($\frac{1}{121}$ + $\frac{1}{122}$ + ... + $\frac{1}{130}$ )

 Ta có:

$\frac{1}{101}$> $\frac{1}{110}$ ;$\frac{1}{102}$ > $\frac{1}{110}$ ; ... ; $\frac{1}{109}$ > $\frac{1}{110}$ ; $\frac{1}{110}$ = $\frac{1}{110}$ .

$\frac{1}{111}$ > $\frac{1}{120}$ ; $\frac{1}{112}$> $\frac{1}{120}$ ; ...; $\frac{1}{119}$ >$\frac{1}{120}$ ; $\frac{1}{120 }$ = $\frac{1}{120}$ 

$\frac{1}{121}$ > $\frac{1}{130}$ ; $\frac{1}{122}$ > $\frac{1}{130}$;... ; $\frac{1}{129}$ > $\frac{1}{130}$ ;$\frac{1}{130}$ = $\frac{1}{130}$ 

Suy ra

S=( $\frac{1}{101}$ + $\frac{1}{102}$ + ... + $\frac{1}{110)}$) + ($\frac{1}{111}$ + $\frac{1}{112}$ + ... + $\frac{1}{120}$ ) + ($\frac{1}{121}$ + $\frac{1}{122}$ + ... + $\frac{1}{130}$ ) > (($\frac{1}{110}$ +$\frac{1}{110}$ + ...+ $\frac{1}{110}$ ) + ($\frac{1}{120}$ + $\frac{1}{120}$ + ...+ $\frac{1}{120}$ ) + ($\frac{1}{130}$ + $\frac{1}{130}$ +...+ $\frac{1}{130}$ )  (10 số hạng $\frac{1}{110}$ ; 10 số hạng $\frac{1}{120}$  ; 10 số hạng $\frac{1}{130}$ )

=> S > ($\frac{1}{110}$ . 10 ) + ($\frac{1}{120}$ . 10) + ($\frac{1}{130}$ . 10 )

      S > $\frac{1}{11}$ + $\frac{1}{12}$ + $\frac{1}{13}$ = $\frac{431}{1716}$ = 0,2511655011... > 0,25 = $\frac{1}{4}$ 

Vậy S>$\frac{1}{4}$ (1)

+)

S= $\frac{1}{101}+$ $\frac{1}{102}$ +...+ $\frac{1}{130}$

   =( $\frac{1}{101}$ + $\frac{1}{102}$ + ... + $\frac{1}{110)}$) + ($\frac{1}{111}$ + $\frac{1}{112}$ + ... + $\frac{1}{120}$ ) + ($\frac{1}{121}$ + $\frac{1}{122}$ + ... + $\frac{1}{130}$ )

 Ta có:

$\frac{1}{101}$< $\frac{1}{100}$ ;$\frac{1}{102}$ < $\frac{1}{100}$ ; ...;$\frac{1}{109}$ < $\frac{1}{100}$ ;$\frac{1}{110}$ < $\frac{1}{100}$ .

$\frac{1}{111}$ < $\frac{1}{110}$ ; $\frac{1}{112}$< $\frac{1}{110}$ ; ...; $\frac{1}{119}$ <$\frac{1}{110}$ ; $\frac{1}{120 }$ < $\frac{1}{110}$ 

$\frac{1}{121}$ < $\frac{1}{120}$ ; $\frac{1}{122}$ < $\frac{1}{120}$;... ; $\frac{1}{129}$ < $\frac{1}{120}$ ; $\frac{1}{130}$ < $\frac{1}{120}$ 

Suy ra

S=( $\frac{1}{101}$ + $\frac{1}{102}$ + ... + $\frac{1}{110)}$) + ($\frac{1}{111}$ + $\frac{1}{112}$ + ... + $\frac{1}{120}$ ) + ($\frac{1}{121}$ + $\frac{1}{122}$ + ... + $\frac{1}{130}$ ) < (($\frac{1}{100}$ +$\frac{1}{100}$ + ...+ $\frac{1}{100}$ ) + ($\frac{1}{110}$ + $\frac{1}{110}$ + ...+ $\frac{1}{110}$ ) + ($\frac{1}{120}$ + $\frac{1}{120}$ +...+ $\frac{1}{120}$ )  (10 số hạng $\frac{1}{100}$ ; 10 số hạng $\frac{1}{110}$  ; 10 số hạng $\frac{1}{120}$ )

=> S < ($\frac{1}{100}$ . 10 ) + ($\frac{1}{110}$ . 10) + ($\frac{1}{120}$ . 10 )

      S < $\frac{1}{10}$ + $\frac{1}{11}$ + $\frac{1}{12}$ = $\frac{181}{660}$ = 0,2742424242... < 0,2757575757... = $\frac{91}{330}$

Vậy S<$\frac{91}{330}$ (2)

Từ (1) và (2) suy ra $\frac{1}{4}$ < S < $\frac{91}{330}$ 

Vậy $\frac{1}{4}$ < S < $\frac{91}{330}$  (dpcm)

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