[lnพwsะบุ๑sสุ๑xล่o]

ข้อความเดิมเขียนโดยคุณ Poomee

$\sum_{n = 1}^{2009} \sqrt{1+\frac{1}{n^2} +\frac{1}{(n+1)^2} }$

$\sum_{n = 1}^{2009} \sqrt{1+\frac{1}{n^2} +\frac{1}{(n+1)^2} }$

$\sum_{n = 1}^{2009} (1+\frac{1}{n} -\frac{1}{(n+1)})$

$=2009+(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{2009})-(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010})$

$=2010-\frac{1}{2010}$