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S có 30 số hạng. Nhóm thành 3 nhóm, mỗi nhóm 10 số hạng
\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{42}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
\(S
Ta có: S = \(\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+...+\frac{1}{60}\right)\)
Nhận xét: \(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{10}{40}=\frac{1}{4}\)
\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}>\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=\frac{10}{50}=\frac{1}{5}\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{10}{60}=\frac{1}{6}\)
\(\Rightarrow S>\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\)
\(\Rightarrow S>\frac{37}{60}>\frac{36}{60}=\frac{3}{5}\) (1)
Lại có: \(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}< \frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)
\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}< \frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{10}{40}=\frac{1}{4}\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}< \frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=\frac{10}{50}=\frac{1}{5}\)
\(\Rightarrow S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}\)
\(\Rightarrow S< \frac{47}{60}< \frac{48}{60}=\frac{4}{5}\) (2)
Từ (1) và (2) => \(\frac{3}{5}< S< \frac{4}{5}\) (đpcm)
AI KẾT BN KO!
TIỆN THỂ TK MÌNH LUÔN NHA!
KONOSUBA!!!
AI TK MÌNH MÌNH TK LẠI 3 LẦN.
\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)
\(S=\left[\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right]+\left[\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right]+\left[\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right]\)
\(S< \left[\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right]+\left[\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right]+\left[\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right]\)
\(S< \frac{10}{30}+\frac{10}{40}+\frac{10}{50}\)
\(S< \frac{37}{60}< \frac{48}{60}=\frac{4}{5}\)
\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+\frac{1}{34}+...+\frac{1}{60}\)\(CMR:S< \frac{4}{5}\)
Số số hạng của S là: (60 - 31 ) + 1 = 30 ( số ), chia thành 6 nhóm, mỗi nhóm 5 số hạng.
Ta có:
\(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+\frac{1}{34}+\frac{1}{35}< \frac{1}{31}+\frac{1}{31}+\frac{1}{31}+\frac{1}{31}+\frac{1}{31}\)
\(\frac{1}{36}+\frac{1}{37}+\frac{1}{38}+\frac{1}{39}+\frac{1}{40}< \frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}\)
\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+\frac{1}{44}+\frac{1}{45}< \frac{1}{41}+\frac{1}{41}+\frac{1}{41}+\frac{1}{41}+\frac{1}{41}\)
\(\frac{1}{46}+\frac{1}{47}+\frac{1}{48}+\frac{1}{49}+\frac{1}{50}< \frac{1}{46}+\frac{1}{46}+\frac{1}{46}+\frac{1}{46}+\frac{1}{46}\)
\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+\frac{1}{55}< \frac{1}{51}+\frac{1}{51}+\frac{1}{51}+\frac{1}{51}+\frac{1}{51}\)
\(\frac{1}{56}+\frac{1}{57}+\frac{1}{58}+\frac{1}{59}+\frac{1}{60}< \frac{1}{56}+\frac{1}{56}+\frac{1}{56}+\frac{1}{56}+\frac{1}{56}\)
\(=>S=\frac{5}{31}+\frac{5}{36}+\frac{5}{41}+\frac{5}{46}+\frac{5}{51}+\frac{5}{56}\)
\(=>S< 0,78...\)\(=>S< \frac{7}{10}\)( mình ước lượng thôi nha )
Vậy \(S< \frac{4}{5}\)vì \(\frac{4}{5}=\frac{8}{10}< \frac{7}{10}\)
~UMK..., mình ko chắc đúng ko nữa~
\(S=\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
Ta có: \(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{10}{40}=\frac{1}{4}\)
\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}>\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=\frac{10}{50}=\frac{1}{5}\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{10}{60}\)
\(\Rightarrow S>\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60}>\frac{36}{60}=\frac{3}{5}\) (1)
Lại có: \(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}< \frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)
\(\frac{1}{41}+...+\frac{1}{50}< \frac{1}{40}+...+\frac{1}{40}=\frac{10}{40}=\frac{1}{4}\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}< \frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=\frac{10}{50}=\frac{1}{5}\)
\(\Rightarrow S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{47}{60}< \frac{48}{60}=\frac{4}{5}\) (2)
Từ (1) và (2) => \(\frac{3}{5}< S< \frac{4}{5}\)