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

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

Có \(1-\frac{1}{46}< 1\)

\(\Rightarrow S< 1\)

nhan xet:3/1.4=1/1-1/4

3/4.7=1/4-1/7

3/7.10=1/7-1/10

.....................

3/40.43=1/40-1/43

3/43.46=1/43-1/46

S=1/1-1/3+1/3-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46

S=1/1-1/46

S=46/46-1/46

S=45/46<1

vay s<1

S=1 - 1/46 < 1 

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 \(S<1\)

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

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

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

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

=> \(\frac{45}{46}<\frac{46}{46}=1\)

=> S<1

S= 1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+|1/43-1/46

S= 1-1/46

S= 45/46<1

vậy S<1

duyệt đi

S=  1- 1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46

S= 1+ (1/4-1/4)+(1/7-1/7)+...+(1/43-1/43)-1/46

S= 1-1/46= 45/46<1

Suy ra S<1

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

= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

= \(1-\frac{1}{46}\)

Vì \(1-\frac{1}{46}< 1\)nên S<1

\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+.......+\frac{3}{43\cdot46}\)

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

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

Ta có \(1-\frac{1}{46}< 1\)=> S < 1

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

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

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

vì \(1-\frac{1}{46}< 1\Rightarrow S< 1\)

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

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

      \(=1-\frac{1}{46}< 1\)

Vậy \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}< 1\)

           Bài làm :

Ta có :

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

\(S=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{43-40}{40.43}+\frac{46-43}{43.46}\)

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

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

=> Điều phải chứng minh

S=1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46

S=1-1/46 S=45/46<1 

vậy S <1 

ta có:3/1.4+3/4.7+....=3/43.46 suy ra 1(1/1-1/4+...+1/43-1/46) suy ra 1(1/1-1/46)= 1.45/46=45/46                                                                         do 45/46<1 suy ra 3/1.4+3/4.7+...+3/43.46<1(DPCM)

\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}+\dfrac{3}{43.46}\\ S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}\\ S=1-\dfrac{1}{46}< 1\)

Vậy S < 1 (đpcm)