Question

Cho \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\).Hãy chứng tỏ rằng S < 1

 

Answer 1

S=1 - 1/46 < 1 

Answer 2

Có \(S=\frac{3}{1.4}+\frac{3}{4.7}+......+\frac{3}{43.46}\)

Sẽ có: \(S=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....+\frac{1}{43.46}\)

\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+........+\frac{1}{43}-\frac{1}{46}\)

\(S=\frac{1}{1}-\left(-\frac{1}{4}+\frac{1}{4}\right)+\left(-\frac{1}{7}+\frac{1}{7}\right)+\left(-\frac{1}{10}+\frac{1}{10}\right)+....+\left(-\frac{1}{43}+\frac{1}{43}\right)-\frac{1}{46}\)

\(S=\frac{1}{1}-0+0+0+0+......+0+0-\frac{1}{46}\)

\(S=\frac{1}{1}-\frac{1}{46}=\frac{45}{46}\)

Vì có:  \(\frac{45}{46}<1\) nên \(S<1\)

Answer 3

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

\(S=1-\frac{1}{46}\)

\(S=\frac{45}{46}\)

=> \(\frac{45}{46}<\frac{46}{46}=1\)

=> S<1