อ้างอิง:
ข้อความเดิมเขียนโดยคุณ กิตติ
3.จงหาค่าของ $\sqrt{1+\frac{4\times 2^2}{(2^2-1)^2} } +\sqrt{1+\frac{4\times 3^2}{(3^2-1)^2} } +\sqrt{1+\frac{4\times 4^2}{(4^2-1)^2} } +...+\sqrt{1+\frac{4\times 20^2}{(20^2-1)^2} } $
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$=\dfrac{2^2+1}{2^2-1}+\cdots +\dfrac{20^2+1}{20^2-1}$
$=\Big(\dfrac{2^2-1+2}{2^2-1}\Big)+\cdots +\Big(\dfrac{20^2-1+2}{20^2-1}\Big)$
$=\Big(1+\dfrac{2}{2^2-1}\Big)+\cdots +\Big(1+\dfrac{2}{20^2-1}\Big)$
$=\Big(1+\dfrac{2}{1\cdot 3}\Big)+\cdots +\Big(1+\dfrac{2}{19\cdot 21}\Big)$
$=\Big(1+\dfrac{1}{1}-\dfrac{1}{3}\Big)+\cdots +\Big(1+\dfrac{1}{19}-\dfrac{1}{21}\Big)$
$=\Big(1+\dfrac{1}{1}+\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{3}\Big)+\cdots +\Big(1+\dfrac{1}{19}+\dfrac{1}{20}-\dfrac{1}{20}-\dfrac{1}{21}\Big)$
$=\Big(1+\dfrac{3}{1\cdot 2}-\dfrac{5}{2\cdot 3}\Big)+\Big(1+\dfrac{5}{2\cdot 3}-\dfrac{7}{3\cdot 4}\Big)+\cdots +\Big(1+\dfrac{39}{19\cdot 20}-\dfrac{41}{20\cdot 21}\Big)$
$=19+\dfrac{3}{1\cdot 2}-\dfrac{41}{20\cdot 21}$
$=\dfrac{41}{2}-\dfrac{41}{20\cdot 21}$
$=\dfrac{8569}{420}$