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

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\(S=\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)

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) => \(\frac{3}{5}< S< \frac{4}{5}\)

Mình trả lời cho 1 bạn rồi đó

ĐÂY NÈ

Các bạn giúp mình nhé

\(S=\left(1+3\right)+...+3^8\left(1+3\right)=4\left(1+...+3^8\right)⋮4\)

\(S=\left(1+3+3^2\right)+...+3^7\left(1+3+3^2\right)\)

\(=13\left(1+...+3^7\right)⋮13\)

\(S=1+3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9\)

\(S=\left(1+3\right)+\left(3^2+3^3\right)+\left(3^4+3^5\right)+\left(3^6+3^7\right)+\left(3^8+3^9\right)\)

\(S=4+3^2\left(1+3\right)+3^4\left(1+3\right)+3^6\left(1+3\right)+3^8\left(1+3\right)\)

\(S=4+3^2.4+3^4.4+3^6.4+3^8.4\)

\(S=4\left(3^2+3^4+3^6+3^8\right)\)

\(4⋮4\\ \Rightarrow4\left(3^2+3^4+3^6+3^8\right)⋮4\\ \Rightarrow S⋮4\)

\(S=3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9\\ =\left(3+3^2+3^3\right)+3^3.\left(3+3^2+3^3\right)+3^6.\left(3+3^2+3^3\right)\\ =39+3^3.39+3^6.39\\ =-39.\left(-1-3^3-3^6\right)⋮\left(-39\right)\)

S = 3 + 3² + 3³ + 3⁴ + 3⁵ + 3⁶ + 3⁷ + 3⁸ + 3⁹

S = ( 3 + 3² + 3³ ) +3⁴ + 3⁵ + 3⁶ + 3⁷ + 3⁸ + 3⁹ 

S = 39 + 3⁴ + 3⁵ + 3⁶ + 3⁷ + 3⁸ + 3⁹

Vì 39 ⋮ -39

<=> S ⋮ -39

\(S=1.\left(1+3\right)+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^8\left(1+3\right)\)

\(S=4x\left(1+3^2+...+3^8\right)\)

Vì 4 chia hết cho 4 nên S chia hết cho 4