วิชาสามัญ | คณิตศาสตร์

$สูตรอนุกรมเลขคณิต a_{n} = a_{1} + (n - 1) d a_{n} = a_{x} + (n - x) d จากโจทย์ a_{1}, a_{2}, a_{3}, \dots, a_{20} เป็นลำดับเลขคณิต ซึ่งมีผลต่างร่วมเท่ากับ \frac{2}{21} แสดงว่า d = \frac{2}{21}$

$จากโจทย์ \frac{1}{21 (a_{20} - a_{1})} + \frac{1}{19 (a_{19} - a_{2})} + \frac{1}{17 (a_{18} - a_{3})} + \dots + \frac{1}{5 (a_{12} - a_{9})} + \frac{1}{3 (a_{11} - a_{10})} \to 1$

$หาค่า a_{20} - a_{1} = (a_{1} + 19 d) - a_{1} = 19 d a_{19} - a_{2} = (a_{2} + 17 d) - a_{2} = 17 d a_{18} - a_{3} = (a_{3} + 15 d) - a_{3} = 15 d a_{12} - a_{9} = (a_{9} + 3 d) - a_{9} = 3 d a_{11} - a_{10} = (a_{10} + d) - a_{10} = d แทนใน 1 จะได้$

$\frac{1}{21 (19 d)} + \frac{1}{19 (17 d)} + \frac{1}{17 (15 d)} + \dots + \frac{1}{5 (3 d)} + \frac{1}{3 (d)} ดึงตัวร่วม \frac{1}{d} แทนใน 1 จะได้; = \frac{1}{d} [\frac{1}{21 (19)} + \frac{1}{19 (17)} + \frac{1}{17 (15)} + \dots + \frac{1}{5 (3)} + \frac{1}{3 (1)}]; คูณ \frac{2}{2} = \frac{2}{2} \times \frac{1}{d} [\frac{1}{19 (21)} + \frac{1}{17 (19)} + \frac{1}{15 (17)} + \dots + \frac{1}{3 (5)} + \frac{1}{1 (3)}] = \frac{1}{2 d} [\frac{2}{19 (21)} + \frac{2}{17 (19)} + \frac{2}{15 (17)} + \dots + \frac{2}{3 (5)} + \frac{2}{1 (3)}]$

$= \frac{1}{2 d} [(\frac{1}{19} - \frac{1}{21}) + (\frac{1}{17} - \frac{1}{19}) + (\frac{1}{15} - \frac{1}{17}) + . . . + (\frac{1}{3} - \frac{1}{5}) + (\frac{1}{1} - \frac{1}{3})]$

$= \frac{1}{2 d} [- \frac{1}{21} + \frac{1}{1}] = \frac{1}{2 d} [1 - \frac{1}{21}] = \frac{1}{2 d} [\frac{20}{21}]; โดย d = \frac{2}{21} = \frac{1}{2 (\frac{2}{21})} [\frac{20}{21}] = 5$

$ดังนั้น ผลรวม \frac{1}{21 (a_{20} - a_{1})} + \frac{1}{19 (a_{19} - a_{2})} + \frac{1}{17 (a_{18} - a_{3})} + \dots + \frac{1}{5 (a_{12} - a_{9})} + \frac{1}{3 (a_{11} - a_{10})} = 5$