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Question

CMR :

$\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+...+ \frac{1}{n^2 +(n+1)^2}$ <

$\frac{1}{2}$ với mọi n thuộc N

congn086 · Answer 1 · 10/28/2017 9:00:03 AM

Ta có:

A =

$\frac{1}{5} + \frac{1}{13} + \frac{1}{25} + ... + \frac{1}{n^2+(n+1)^2}$

A =

$\frac{1}{1^2+2^2} + \frac{1}{2^2+3^2} + \frac{1}{3^2+4^2} + ... + \frac{1}{n^2+(n+1)^2}$

Lại có:

$n^2 + (n+1)^2 > 2n(n+1)$ ( BĐT Cô-si và n < n+1 )

=>

$A < \frac{1}{2.1.2} + \frac{1}{2.2.3} + \frac{1}{2.3.4} + ... +\frac{1}{2.n.(n+1)} )$

$A < \frac{1}{2}\times ( \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + ... + \frac{1}{n.(n+1)})$

$A < \frac{1}{2}\times (1-\frac{1}{2} + \frac{1}{2} - \frac{1}{3} +\frac{1}{3} - \frac{1}{4} + ... + \frac{1}{n} - \frac{1}{n+1})$

$A < \frac{1}{2}\times ( 1 - \frac{1}{n+1}) < \frac{1}{2}$

Sửa 28-10-17 09:00 AM

congn086
61 5

94K 66K

Trả lời 28-10-17 08:57 AM

congn086
61 5

94K 66K

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