[tongkub]

รบกวนช่วยเช็คด้วยนะครับ

ข้อ 2

$A = \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \frac{1}{8^2} + .....$

$B = \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + ..... $

ให้ $S = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + .....$


$A = \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \frac{1}{8^2} + .....$

$A = (\frac{1}{2^2})(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + .....)$

$A = \frac{S}{4}$

$S = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + .....$

$ S - A = \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + ..... = B$

$\therefore B = S - A = S - \frac{S}{4} = \frac{3S}{4}$

$\frac{A}{B} = \dfrac{\dfrac{A}{4}}{\dfrac{3A}{4}} = \frac{1}{3}$

ข้อ4

$S.D. = \sqrt{ \dfrac{R^2}{N} - (\bar R )^2}$

จาก พื้นที่เฉลี่ย = $12\pi$

$\pi{\sum_{i = 1}^{100}\dfrac{(R_i^2)}{100}} = 12\pi$

$\sum_{i = 1}^{100}R^2 = 1200$


$S.D. = \sqrt{ \dfrac{1200}{100} - (3 )^2}$

= $\sqrt{3}$