เส้นทแยงมุม $D_n= \dfrac{n(n-3)}{2}$
$\dfrac{1}{D_n} = \dfrac{2}{n(n-3)}$
$\dfrac{2}{D_4} + \dfrac{2}{D_5} + \dfrac{2}{D_6} +\dfrac{2}{D_7} + ... + \dfrac{2}{D_{2012}}$
$= 2 \left[\dfrac{1}{D_4} + \dfrac{1}{D_5} + \dfrac{1}{D_6} +\dfrac{1}{D_7} + ... + \dfrac{1}{D_{2012}} \right]$
$= 2 \left[\dfrac{1}{4(4-3)} + \dfrac{1}{5(5-3)} + \dfrac{1}{6(6-3)} +\dfrac{1}{7(7-3)} + ... + \dfrac{1}{2012(2012-3)} \right]$
$= 2 \left[\dfrac{1}{(1\times 4)} + \dfrac{1}{(2\times 5)} + \dfrac{1}{(3\times 6)} +\dfrac{1}{(4\times 7)} + ... + \dfrac{1}{(2009 \times 2012)} \right]$
$= \dfrac{2}{3} \left[\dfrac{3}{(1\times 4)} + \dfrac{3}{(2\times 5)} + \dfrac{3}{(3\times 6)} +\dfrac{3}{(4\times 7)} + ... + \dfrac{3}{(2009 \times 2012)} \right]$
$= \dfrac{2}{3} \left[ (\dfrac{1}{1} - \dfrac{1}{4}) + (\dfrac{1}{2} - \dfrac{1}{5}) + (\dfrac{1}{3} - \dfrac{1}{6}) + (\dfrac{1}{4} - \dfrac{1}{7}) + ... + (\dfrac{1}{2009} - \dfrac{1}{2012})\right]$
$= \dfrac{2}{3} \left[ \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{2010} - \dfrac{1}{2011} - \dfrac{1}{2012}\right]$
ทำไมตัวเลขแยะจัง ผิดตรงไหนหรือเปล่า ?
มาอ่านโจทย์ใหม่
$D_n \ $ไม่ใช่เส้นทแยงมุม แต่เป็น ผลรวมเส้นทแยงมุมกับจำนวนด้าน
$D_n \ = 1 + 2 +3 + ... + (n-1)$
$D_n \ = \dfrac{n(n-1)}{2}$
$\dfrac{1}{D_n} = \dfrac{2}{n(n-1)}$
$ \dfrac{1}{D_4} + \dfrac{1}{D_5} + \dfrac{1}{D_6} +\dfrac{1}{D_7} + ... + \dfrac{1}{D_{2012}} $
$= 2 \left[\dfrac{1}{(4)(4-1)} + \dfrac{1}{(5)(5-1)} + \dfrac{1}{(6)(6-1)} +\dfrac{1}{(7)(7-1)} + ... + \dfrac{1}{(2012)(2012-1)} \right]$
$= 2 \left[\dfrac{1}{(4)(3)} + \dfrac{1}{(5)(4)} + \dfrac{1}{(6)(5)} +\dfrac{1}{(7)(6)} + ... + \dfrac{1}{(2012)(2011)} \right]$
$= 2 \left[(\dfrac{1}{3} - \dfrac{1}{4}) + (\dfrac{1}{4} - \dfrac{1}{5}) +(\dfrac{1}{5} - \dfrac{1}{6}) + ... + (\dfrac{1}{2011} - \dfrac{1}{2012}) \right]$
$ = 2 (\dfrac{1}{3} - \dfrac{1}{2012})$
$ = \dfrac{2009}{3018} = \dfrac{a}{b}$
$b - a = 1009$
ตอบ ข้อ ก.