[แมวสามสี]

ข้อ 21. ให้ $A=\bmatrix{a & b \\ c & d} $ จะำได้$\vmatrix{a & b \\ c & d} =ad-bc = 4$........(1)

ทีนี้ก็ใช้$\left|A-3I\,\right| =0$ ได้ $ (a-3)(d-3)-bc=0$ ลองต่อดูครับ

ข้อ31. จาก $\sum_{n = 2}^{\infty} \frac{1}{n^4-n^2} =A$

พิจารณา$\sum_{n = 2}^{\infty} \frac{1}{n^4-n^2} =\sum_{n = 2}^{\infty} \frac{1}{(n^2-1)n^2}$

$\qquad\qquad\qquad\qquad=\frac{1}{3\times 4}+\frac{1}{8\times 9}+ \frac{1}{15\times 16}+\frac{1}{24\times 25}+\frac{1}{35\times 36}+ \frac{1}{48\times 49}+ ...$

$\qquad\qquad\qquad\qquad=\frac{1}{3} -\frac{1}{4} +\frac{1}{8} -\frac{1}{9} +\frac{1}{15} -\frac{1}{16}+\frac{1}{24} -\frac{1}{25}+\frac{1}{35} -\frac{1}{36}+\frac{1}{48} -\frac{1}{49}+...$

$\qquad\qquad\qquad\qquad=( \frac{1}{3}+\frac{1}{8}+\frac{1}{15}+\frac{1}{24}+\frac{1}{35}+\frac{1}{48}+...)-( \frac{1}{4} +\frac{1}{9} +\frac{1}{16}+\frac{1}{25}+\frac{1}{36}+\frac{1}{49}+...)$

$\qquad\qquad\qquad\qquad=( \frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...) +(\frac{1}{8}+\frac{1}{24}++\frac{1}{48}+...)-\sum_{n = 2}^{\infty} \frac{1}{n^2}$

$\qquad\qquad\qquad\qquad=(\frac{1}{1\times 3}+\frac{1}{3\times 5}+ \frac{1}{5\times 7}+...)+(\frac{1}{2\times 4}+\frac{1}{4\times 6}+ \frac{1}{6\times 8}+...)-\sum_{n = 2}^{\infty} \frac{1}{n^2}$

$\qquad\qquad\qquad\qquad=\frac{1}{2} (\frac{1}{1} -\frac{1}{3} +\frac{1}{3} -\frac{1}{5} +\frac{1}{5} -\frac{1}{7}+...)+\frac{1}{2} (\frac{1}{2} -\frac{1}{4} +\frac{1}{4} -\frac{1}{6} +\frac{1}{6} -\frac{1}{8}+...)-\sum_{n = 2}^{\infty} \frac{1}{n^2}$

$\qquad\qquad\qquad\qquad=\frac{1}{2}+\frac{1}{4}-\sum_{n = 2}^{\infty} \frac{1}{n^2}$

$\qquad\qquad\qquad\qquad=\frac{3}{4}-\sum_{n = 2}^{\infty} \frac{1}{n^2}$

ได้$ \qquad A=\frac{3}{4}-\sum_{n = 2}^{\infty} \frac{1}{n^2}$

ดังนั้น $ \qquad \sum_{n = 2}^{\infty} \frac{1}{n^2} = \frac{3}{4}-A$