Cho tổng S = 1/11 + 1/12 + 1/13 +...+1/40, chứng tỏ rằng 1< S <2, từ đó suy ra S không phải là số tự nhiên
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Ta có :
S = 1 + 3 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 310 + 311
S = (1 + 3 + 32) + (33 + 34 + 35 ) + (36 + 37 + 38 ) + (39 + 310 + 311)
S = 1 . (1 + 3 + 32 ) + 33 . (1 + 3 + 32) + 36 . (1 + 3 + 32) + 39 . (1 + 3 + 32)
S = 1 . 13 + 33 . 13 + 36 . 13 + 39 . 13
S = 13 . (1 + 33 + 36 + 39) chia hết cho 13 nên S chia hết cho 13 (ĐPCM)
S = 1 + 3 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 310 + 311
S = (1 + 3 + 32 + 33) + (34 + 35 + 36 + 37) + (38 + 39 + 310 + 311)
S = 1.(1 + 3 + 32 + 33) + 34 . (1 + 3 + 32 + 33) + 38 . (1 + 3 + 32 + 33)
S = 1 . 40 + 34 . 40 + 38 .40
S = 40 . (1 + 34 + 38) chia hết cho 40 (ĐPCM)
Ủng hộ mk nha !!! ^_^
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a: Ta có
A = \(\dfrac{1}{10}\) + \((\dfrac{1}{11}\) + \(\dfrac{1}{12}\) + ...+ \(\dfrac{1}{100}\)\()\)
⇒ A > \(\dfrac{1}{10}\) + \((\dfrac{1}{100}\) + \(\dfrac{1}{100}\) + ...+ \(\dfrac{1}{100}\)\()\)90 số hạng
⇒ A > \(\dfrac{1}{10}\) + \(\dfrac{90}{100}\)
⇒ A > 1
vậy A > 1
b: ta có
S = (\(\dfrac{1}{21}\) + \(\dfrac{1}{22}\)+ \(\dfrac{1}{23}\) + \(\dfrac{1}{24}\) + \(\dfrac{1}{25}\))+(\(\dfrac{1}{26}\) + \(\dfrac{1}{27}\)+ \(\dfrac{1}{28}\) + \(\dfrac{1}{29}\) + \(\dfrac{1}{30}\))+(\(\dfrac{1}{31}\) + \(\dfrac{1}{32}\)+ \(\dfrac{1}{33}\) + \(\dfrac{1}{34}\) + \(\dfrac{1}{35}\))
⇒ S > (\(\dfrac{1}{25}\) + \(\dfrac{1}{25}\)+ \(\dfrac{1}{25}\) + \(\dfrac{1}{25}\) + \(\dfrac{1}{25}\))+(\(\dfrac{1}{30}\) + \(\dfrac{1}{30}\)+ \(\dfrac{1}{30}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{30}\))+(\(\dfrac{1}{35}\) + \(\dfrac{1}{35}\)+ \(\dfrac{1}{35}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{35}\))
⇔ S > \(\dfrac{5}{25}\)+\(\dfrac{5}{30}\)+\(\dfrac{5}{35}\)
⇔ S > \(\dfrac{1}{5}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{7}\)
⇔ S > \(\dfrac{107}{210}\)> \(\dfrac{105}{210}\)=\(\dfrac{1}{2}\)
vậy S > \(\dfrac{1}{2}\)
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Ta có: 1/20<1/11
1/20<1/12
...
=> 1/20+1/20+..+1/20 < 1/11+1/12+...+1/20
=> 1/20.10<1/11.1/12+1/13+...+1/20
=> 1/2< 1/11+1/12+1/12+1/13+...+1/20
=> 1/2<S (đpcm)
k mik nhé các bạn. Thanks you nhé ^_<