ข้อสอบ PAT 1 - เมษายน 2557 | ความถนัดทางคณิตศาสตร์

$จากโจทย์ a_{n} = \frac{n^{2}}{16 n^{2} - 4} \to 1 พิจารณา \frac{n^{2}}{16 n^{2} - 4} จะได้ \frac{n^{2}}{16 n^{2} - 4} = \frac{n^{2}}{4 (4 n^{2} - 1)} = \frac{n^{2}}{4 [{(2 n^{2})}^{2} - 1^{2}]} = \frac{n^{2}}{4 (2 n - 1) (2 n + 1)} = \frac{n^{2}}{4} (\frac{1}{(2 n - 1) (2 n + 1)})$

$= \frac{n^{2}}{4} (\frac{1}{2} (\frac{1}{2 n - 1} - \frac{1}{2 n + 1})) = \frac{1}{8} (\frac{n^{2}}{2 n - 1} - \frac{n^{2}}{2 n + 1}) แสดงว่า a_{n} = \frac{1}{8} (\frac{n^{2}}{2 n - 1} - \frac{n^{2}}{2 n + 1})$

$จากโจทย์ \lim_{n \to \infty} \frac{a_{1} + a_{2} + a_{3} + . . . + a_{n}}{n} = \frac{a}{b} \to 2 พิจารณา a_{1} + a_{2} + a_{3} + . . . + a_{n} จะได้ a_{1} + a_{2} + a_{3} + . . . + a_{n}$

$= \frac{1}{8} [(\frac{1^{2}}{1} - \frac{1^{2}}{3}) + (\frac{2^{2}}{3} - \frac{2^{2}}{5}) + (\frac{3^{2}}{5} - \frac{3^{2}}{7}) + . . . + (\frac{n^{2}}{2 n - 1} - \frac{n^{2}}{2 n + 1})] = \frac{1}{8} [\frac{1^{2}}{1} + (\frac{2^{2}}{3} - \frac{1^{2}}{3}) + (\frac{3^{2}}{5} - \frac{2^{2}}{5}) + . . . + (\frac{n^{2}}{2 n - 1} - \frac{{(n - 1)}^{2}}{2 n - 1}) - \frac{n^{2}}{2 n + 1}]$
$= \frac{1}{8} [\frac{1^{2}}{1} + (\frac{(2 - 1) (2 + 1)}{3}) + (\frac{(3 - 2) (3 + 2)}{5}) + . . . + (\frac{[n - (n - 1)] [n + (n - 1)]}{2 n - 1}) - \frac{n^{2}}{2 n + 1}]$
$= \frac{1}{8} [\underset{n}{\underset{⏟}{1 + 1 + 1 + . . . + 1}} - \frac{n^{2}}{2 n + 1}] = \frac{1}{8} [n - \frac{n^{2}}{2 n + 1}]$

$แสดงว่า a_{1} + a_{2} + a_{3} + . . . + a_{n} = \frac{1}{8} [n - \frac{n^{2}}{2 n + 1}] แทนใน 2 จาก 2 จะได้ \lim_{n \to \infty} \frac{\frac{1}{8} [n - \frac{n^{2}}{2 n + 1}]}{n} = \frac{a}{b} \lim_{n \to \infty} \frac{1}{8 n} [n - \frac{n^{2}}{2 n + 1}] = \frac{a}{b} \lim_{n \to \infty} \frac{1}{8} [1 - \frac{n}{2 n + 1}] = \frac{a}{b} \frac{1}{8} (1 - \frac{1}{2}) = \frac{a}{b} \frac{1}{16} = \frac{a}{b}$