\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3n}{n\left(n+3\right)}\)
\(S=\frac{4-1}{1\cdot4}+\frac{7-4}{4\cdot7}+\frac{10-7}{7\cdot10}+...+\frac{n+3-3}{n\left(n+3\right)}\)
\(S=\frac{4}{1\cdot4}-\frac{1}{1\cdot4}+\frac{7}{4\cdot7}-\frac{4}{4\cdot7}+...+\frac{n+3}{n\left(n+3\right)}-\frac{n}{n\left(n+3\right)}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n-3}\)
\(S=1-\frac{1}{n-3}=\frac{n-3}{n-3}-\frac{1}{n-3}=\frac{n-3-1}{n-3}=\frac{n-2}{n-3}< 1\)
bé hơn 1 chứ ko lớn hơn 1 đc đâu
cute phô mai que chắc mình nhầm đề. Thanks bn nha <3
\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{n\left(n+3\right)}\)
\(\Rightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n-3}\)
\(\Rightarrow S=1-\frac{1}{n+3}\)
\(\Rightarrow S=\frac{n+2}{n+3}\)
\(\Rightarrow S< 1\)
chị CTV sai kìa phần cuối \(\frac{n-2}{n-3}\)thì vì 2 < 3 nên n - 2 > n - 3 nên S > 1 chứ :)
