[กิตติ]

ข้อ 12 ตอน2
$\frac{a+b+c+...+x+y}{a+b+c+...+x+y+z} =\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{25}} $

$S=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{25}} $......(1)
$2S =1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{24}}$........(2)
(2)-(1) $S =1-\frac{1}{2^{25}}=\frac{2^{25}-1}{2^{25}} $

โจทย์ถาม $\frac{a+b+c+...+x+y+z}{z} $

$\frac{a+b+c+...+x+y}{a+b+c+...+x+y+z}+\frac{z}{a+b+c+...+x+y+z} =1$
$\frac{z}{a+b+c+...+x+y+z} =1-\frac{a+b+c+...+x+y}{a+b+c+...+x+y+z}=1-\left(\,\frac{2^{25}-1}{2^{25}}\right) $
$\frac{z}{a+b+c+...+x+y+z} =\frac{1}{2^{25}}$

$\frac{a+b+c+...+x+y+z}{z}= 2^{25}$