Question

[ngô trà my]

Cho s = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n\left(n+3\right)}\)n \(\in\)N KHÁC 0. Chứng minh s < 1

Answer 1

[Trần Lan Phương]

• 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 = \(1-\frac{1}{n+3}\)

\(\Rightarrow\) S < 1 ( đpcm )

Answer 2

[Sesshomaru]

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

=> S = 1 - \(\frac{1}{n+3}\)

vậy S = 1-  \(\frac{1}{n+3}\)