Question

Cho: S=\(\frac{3}{1x4}+\frac{3}{4x7}+\frac{3}{7x10}+...+\frac{3}{100x103}\). Chứng minh S<1

Answer 1

S=1/1-1/4+1/4+1/7-1/7+1/10+...+1/100-1/103

S=1/1-1/103

S=102/103

Vì 102/103<1 nên S<1

Answer 2

\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{100\cdot103}\)

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\)

\(S=1-\frac{1}{103}\)

\(S=\frac{102}{103}< 1\)

Answer 3

\(\frac{3}{1x4}+\frac{3}{4x7}+\frac{3}{7x10}+.......+\frac{3}{100x103}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}....+\frac{1}{100}-\frac{1}{103}\)

\(=\frac{1}{1}-\frac{1}{103}\)

=\(\frac{102}{103}\)

Answer 4

suy ra :S<1

mình thiếu cái này trong bài nha

Answer 5

\(\Rightarrow S=\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{103.106}\)<1

\(\Rightarrow S=3.\left(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{103.106}\right)< 1\)

\(\Rightarrow S=3.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{103}-\frac{1}{106}\right)\)< 1

\(\Rightarrow S=3.\left(\frac{1}{4}-\frac{1}{106}\right)\)< 1

\(\Rightarrow S=3.\frac{51}{212}\)< 1

\(\Rightarrow S=\frac{153}{212}< 1\)

Vì \(\frac{153}{212}< 1\Rightarrow S< 1\)

Answer 6

S = 3/1×4 + 3/4×7 + 3/7×10 + ... + 3/100×103 

S = 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + .... + 1/100 - 1/103

S = 1 - 1/103

S = 103/103 - 1/103

S = 102/103 < 1 

Vậy S < 1 ( điều phải chứng minh ) 

Answer 7

S = 3/1×4 + 3/4×7 + 3/7×10 + ... + 3/100×103

S = 1 ‐ 1/4 + 1/4 ‐ 1/7 + 1/7 ‐ 1/10 + .... + 1/100 ‐ 1/103

S = 1 ‐ 1/103 S = 103/103 ‐ 1/103

S = 102/103 < 1

Vậy S < 1