вычеслите: а)3+1+1/3+...; б)1/2+1/4+1/8+... в)8-4+2-...; г)1-3/4+9/16.... Created by olechka199812. algebra-ru

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olechka199812 @olechka199812

December 2021 1 21 Report

вычеслите: а)3+1+1/3+...; б)1/2+1/4+1/8+... в)8-4+2-...; г)1-3/4+9/16...

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Матов

Verified answer

Это убывающая геометрическая прогрессия
$S=\frac{b_{1}}{1-q}$
$a)\\
b_{1}=3\\\
q=\frac{1}{3}\\\ 
\frac{3}{1-\frac{1}{3}}=\frac{3}{\frac{2}{3}}=4.5$

$b)\\\
b_{1}=\frac{1}{2}\\
q=\frac{1}{2}\\
S=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1$

$c)\\\
8-4+2-1+\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}...=4+1+\frac{1}{4}+\frac{1}{16}..\\
b_{1}=4\\
q=\frac{1}{4}\\
S=\frac{4}{1-\frac{1}{4}} = \frac{4}{\frac{3}{4}}=\frac{16}{3}$

$d)\\\
 (1-\frac{3}{4})+(\frac{9}{16}-\frac{27}{64})+(\frac{81}{256}-\frac{243}{1024})...\\\
\frac{1}{4}+\frac{9}{64}+\frac{81}{1024}\\
b_{1}=\frac{1}{4}\\\
q=\frac{9}{16}\\\
S=\frac{\frac{1}{4}}{1-\frac{9}{16}} = \frac{4}{7}$

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