Question

Cho S = 3/1.4 +3/4.7 +3/7.10 +...+3/n(n+3).

n thuộc số tự nhiên khác 0

Chứng minh S < 1

Làm nhanh lên hộ mk nhé

Giari chi tiết giùm mk nhé

 

Answer 1

S = 3/1 . 4 + 3/4 . 7 + 3/7 . 10 + ...+ 3/n . ( n + 3 ) 

S = 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ...+ 1/n - 1/n + 3 

S = 1 - 1/n + 3  < 1 

S < 1 ( Đpcm ) 

Tham khảo nha !!! 

Answer 2

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

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

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

Vậy S < 1

Answer 3

\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{n\left(n+3\right)}\)

\(S=\frac{4-1}{1\cdot4}+\frac{7-4}{4\cdot7}+...+\frac{n+3-n}{n\left(n+3\right)}\)

\(S=\frac{4}{1.4}-\frac{1}{1.4}+\frac{7}{4.7}-\frac{4}{4.7}+...+\frac{n+3}{n\left(n+3\right)}-\frac{n}{n\left(n+3\right)}\)

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

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

\(\Rightarrow S< 1\)

Answer 4

\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{n\left(n+3\right)}\)

\(S=\frac{4-1}{1\cdot4}+\frac{7-4}{4\cdot7}+\frac{10-7}{7\cdot10}+...+\frac{\left(n+3\right)-n}{n\left(n+3\right)}\)

\(S=\left(\frac{4}{1\cdot4}-\frac{1}{1\cdot4}\right)+\left(\frac{7}{4\cdot7}-\frac{4}{4\cdot7}\right)+...+\left(\frac{n+3}{n\left(n+3\right)}-\frac{n}{n\left(n+3\right)}\right)\)

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

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

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