Cho \(S=\) \(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+........+\frac{1}{60}\) CMR: \(\frac{3}{5}< S< \frac{4}{5}\)
Cho \(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+\frac{1}{34}+...+\frac{1}{60}\)
CMR : \(\frac{3}{5}< S< \frac{4}{5}\)
S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+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 + ...+ 1/50) > 1/5 ; (1/51 + 1/52+...+1/59+1/60) > 1/6
S > 1/4 + 1/5 + 1/6.
Trong khi đó (1/4 + 1/5 + 1/6) > 3/5
Vậy A > 3/5
Phần 2.
S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+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 + ...+ 1/50) < 1/4 ; (1/51 + 1/52+...+1/59+1/60) < 1/5
Mà S = (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 S < 4/5
cho S=\(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\) c/m \(\frac{3}{5}< S< \frac{4}{5}\)
\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)
\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{49}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
\(\frac{1}{40}.10+\frac{1}{50}.10+\frac{1}{60}.10< S< \frac{1}{30}.10+\frac{1}{40}.10+\frac{1}{50.10}\)
\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}< S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}\)
\(\frac{1}{4}+\frac{1}{5}+\frac{3}{20}< \frac{1}{4}+\frac{1}{5}+\frac{1}{6}< S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}< \frac{1}{3}+\frac{4}{15}+\frac{1}{5}\)
\(\frac{3}{5}< S< \frac{4}{5}\left(đpcm\right)\)
Cho S = \(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\) . Chứng minh rằng \(\frac{3}{5}< S< \frac{4}{5}\).
a) Cho \(s=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)
CMR 1<s<2, từ đó suy ra s ko phải stn
b) Cho \(s=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+....+\frac{1}{60}\)
CMR 3/5< s < 4/5
a)ta có:
\(\frac{3}{10}\)>\(\frac{3}{15}\)
\(\frac{3}{11}\)>\(\frac{3}{15}\)
...
\(\frac{3}{14}\)>\(\frac{3}{15}\)
Cộng từng vế của bất đẳng thức trên ta được:
\(\frac{3}{10}\)+\(\frac{3}{11}\)+\(\frac{3}{12}\)+\(\frac{3}{13}\)+\(\frac{3}{14}\)<\(\frac{3}{15}\)+\(\frac{3}{15}\)+\(\frac{3}{15}\)+\(\frac{3}{15}\)+\(\frac{3}{15}\)
Hay S>\(\frac{15}{15}\)=>S>1 (1)
ta có :
\(\frac{3}{11}\)<\(\frac{3}{10}\)
\(\frac{3}{12}\)<\(\frac{3}{10}\)
...
\(\frac{3}{14}\)<\(\frac{3}{10}\)
Cộng từng vế của bất đẳng thức trên ta được:
\(\frac{3}{10}\)+\(\frac{3}{11}\)+\(\frac{3}{12}\)+\(\frac{3}{13}\)+\(\frac{3}{14}\)<\(\frac{3}{10}\)+\(\frac{3}{10}\)+\(\frac{3}{10}\)+\(\frac{3}{10}\)+\(\frac{3}{10}\)
Hay S<\(\frac{15}{10}\)<\(\frac{20}{10}\)=2
Vậy S<2 (2)
Theo câu 1 ta có : S>1
Theo câu 2 ta có :S<2
Vậy 1<S<2
=>S ko phải số tự nhiên
Cho S =\(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)
Chứng Minh Rằng \(\frac{3}{5}
Cho S=\(\)\(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+......+\frac{1}{60}\)
CMR:\(\frac{3}{5}\)<S<\(\frac{4}{5}\)
\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)
So sánh S và 4/5
ớ chết, mk nhầm, lm lại nha
\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)
\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
\(S< \frac{1}{30}.10+\frac{1}{40}.10+\frac{1}{50}.10\)
\(S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}< \frac{4}{5}\)
=> \(S< \frac{4}{5}\)
\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)
\(S< 30.\frac{1}{60}\)
\(S< \frac{1}{2}< \frac{4}{5}\)
\(S< \frac{4}{5}\)
\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)
\(=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
\(\Rightarrow S< \frac{1}{30}.10+\frac{1}{40}.10+\frac{1}{50}.10\)
\(S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}< \frac{4}{5}\)
\(V\text{ậy}:S< \frac{4}{5}\)
Chung minh rang:\(\frac{3}{5}< S< \frac{4}{5}\)
S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+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 + ...+ 1/50) > 1/5 ; (1/51 + 1/52+...+1/59+1/60) > 1/6
S > 1/4 + 1/5 + 1/6.
Trong khi đó (1/4 + 1/5 + 1/6) > 3/5
=>S > 3/5 (1)
S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)
Mà : (1/31+1/32+1/33+...+1/40) < 1/31 x 10 = 10/30 = 1/3 (gồm 10 số hạng)
=> S < 4/5 (2)
Từ (1) và (2) => 3/5 <S<4/5
Cho S =\(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)Chứng minh rằng \(\frac{3}{5}\)<S<\(\frac{4}{5}\)