ตอบพี่ข้างบนนะครับ
$\frac{1}{4}log2+\frac{1}{8}log4+\frac{1}{16}log8+\frac{1}{32}log16+\frac{1}{64}log32+...$
$=\frac{1}{4}log2+\frac{1}{8}log2^2+\frac{1}{16}log2^3+\frac{1}{32}log2^4+\frac{1}{64}log2^5+...$
$S_\infty =\frac{1}{4}log2+\frac{2}{8}log2+\frac{3}{16}log2+\frac{4}{32}log2+\frac{5}{64}log2+...$
$S_\infty =(\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+\frac{5}{64}+...)log2$___________(1)
$(1)\times r;(1) \times \frac{1}{2}$
$\frac{1}{2}S_\infty =(\frac{1}{8}+\frac{2}{16}+\frac{3}{32}+\frac{4}{64}+\frac{5}{128}+...)log2$__________(2)
(1)-(2)
$\frac{1}{2}S_\infty=(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+...)log2$
$\frac{1}{2}S_\infty=(\frac{a_1}{1-r})log2$
$\frac{1}{2}S_\infty=(\frac{\frac{1}{4}}{1-\frac{1}{2}})log2$
$\frac{1}{2}S_\infty=\frac{1}{2}log2$
$S_\infty=log2$
