Question

Cho S = \(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+\(\frac{3}{7.10}\)+ . . . + \(\frac{3}{n\left(n+3\right)}\)( n thuộc N^* )

Chứng minh S <1

Answer 1

\(S=\frac{3}{1.4}+\frac{3}{4.7}+......+\frac{3}{n\left(n+3\right)}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{n}-\frac{1}{n+3}\)

\(=1-\frac{1}{n+3}\)

Ta có :

\(\frac{1}{n+3}>0\)

\(\Leftrightarrow-\frac{1}{n+3}< 0\)

\(\Leftrightarrow1-\frac{1}{n+3}< 1\)

\(\Leftrightarrow S< 1\left(đpcm\right)\)

Answer 2

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n.\left(n+3\right)}\)

 \(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)

\(S=1-\frac{1}{n+3}\)

\(S=\frac{n+2}{n+3}\)

Vi \(n\inℕ^∗\)nên \(n+2< n+3\)

DO đó\(\frac{n+2}{n+3}< 1\)

Vậy S <1

Answer 3

\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+3}\)

\(=1-\frac{1}{n+3}< 1\)

Answer 4

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+3}\)

\(=1-\frac{1}{n+3}\)

=\(\frac{n+3}{n+3}-\frac{1}{n+3}=\frac{n+2}{n+3}< 1\)