Ta có:
\(S=\dfrac{3}{3}.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{43}-\dfrac{1}{46}\right)\)
\(S=1.\left(\dfrac{1}{1}-\dfrac{1}{46}\right)\)
\(S=1.\dfrac{45}{46}=\dfrac{45}{46}\)
Vì \(\dfrac{45}{46}< \dfrac{46}{46}\) nên \(\dfrac{45}{46}< 1\).
Vậy S < 1.
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{43.46}\)
\(S=\dfrac{3}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{43.46}\right)\)
Ta thấy:
\(\dfrac{3}{1.4}=1-\dfrac{1}{4};\dfrac{3}{4.7}=\dfrac{1}{4}-\dfrac{1}{7};\dfrac{3}{7.10}=\dfrac{1}{7}-\dfrac{1}{10};\)
\(...;\dfrac{3}{43.46}=\dfrac{1}{43}-\dfrac{1}{46}\)
\(\Rightarrow S=1\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{43}-\dfrac{1}{46}\right)\)
\(\Rightarrow S=1\left(1-\dfrac{1}{46}\right)\)
\(\Rightarrow S=1.\dfrac{45}{46}=\dfrac{45}{46}\)
Ta có S=1-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{10}\)+...+\(\dfrac{1}{43}\)-\(\dfrac{1}{46}\)
=1-\(\dfrac{1}{46}\)<1
=>S<1
Mk góp ý nha! Bn nên ghi "cmr" ra cụ thể ik dù bít đó là "chứng minh rằng" hưng có 1 số ng` k hỉu lại ns lak bn ns bậy!
S=1-1/4+1/4-1/7+1/7-1/10+...+1/43-1/46
S=(1-1/46)+(1/4-1/4)+...+(1/43-1/43)
S=(46/46-1/46)+0+...+0
S=45/46
Mà 45/46<46/46
Suy ra:45/46<1
Suy ra:S<1
Vậy S<1.
Ta có: \(S=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{43\cdot46}\)
\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{43}-\dfrac{1}{46}\)
\(S=1-\dfrac{1}{46}< 1\)
Vậy \(S< 1\)
