\(S=\left(1+2\right)+...+2^6\left(1+2\right)=3\left(1+...+2^6\right)⋮3\)

\(S=\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+...+\frac{5}{49}\)

\(S>5\left(\frac{1}{49}+\frac{1}{49}+...+\frac{1}{49}\right)\)(30 số hạng \(\frac{1}{49}\))

\(\Leftrightarrow S>5.\frac{30}{49}\)

\(\Leftrightarrow S>\frac{150}{49}=3\frac{3}{49}\)

\(\Rightarrow S>3\)

\(\Rightarrow S>\frac{3}{49}\)

Vậy \(3< S\)  (1)

Ta lại có: \(S< 5.\left(\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\right)\)(30 số hạng)

\(S< \frac{30}{20}.5=\frac{150}{20}=\frac{15}{2}=7\frac{1}{2}\)

\(\Rightarrow S< 7< 8\)

\(\Rightarrow S< \frac{1}{2}\)

Vậy \(S< 8\) (2)

Từ (1) và (2) ta có đpcm

thank you very much

khi minh vua dang

S = (1+ 2)+(22 + 23 )+( 24 + 27) + (26 + 25)

S=   3+45+51+51

S=3+3.15+3.17+3.17

S=3.(1+15+17.2): hết 3

tick nha nhanh nhất nè

S=(1+2)+...+2^6(1+2)=3(1+...+2^6)⋮3

a: \(A=2\left(1+2+2^2\right)+...+2^{19}\left(1+2+2^2\right)\)

\(=7\left(2+...+2^{19}\right)⋮7\)

tự làm nha

a: \(A=2\left(1+2+2^2\right)+...+2^{19}\left(1+2+2^2\right)\)

\(=7\left(2+...+2^{19}\right)⋮7\)

a: \(A=2\left(1+2+2^2\right)+...+2^{19}\left(1+2+2^2\right)\)

\(=7\cdot\left(2+...+2^{19}\right)⋮7\)