Prosze o rozwiazenie 2 podpunktow ciągi geometryczne oblicz sumy c) 16+ 8+ 4+ 2+ ... + 1/32d) 3/4+ 1.... Question from @ZeW95

Roma $(a_n)$ - ciąg geometryczny
$a_1, \ a_2, \ \dots \ a_n$ - kolejne wyrazy ciągu
$q$ - iloraz ciągu geometrycznego
$S_n$ - suma n początkowych wyrazów ciągu

c)
$16+ 8+ 4+ 2+ \dots + \frac{1}{32} \\\\
a_1 = 16, \ a_2 =8 \\\\ 
q = a_2 : a_1 = 8 : 16= \frac{8}{16} =\frac{1}{2} \\\\
a_n = a_1 \cdot q^{n-1}$

$16 \cdot (\frac{1}{2})^{n-1} = \frac{1}{32} \ / :16 \\\\
(\frac{1}{2})^{n-1} = \frac{1}{32} \cdot \frac{1}{16} \\\\
(\frac{1}{2})^{n-1} = \frac{1}{32 \cdot 16} \\\\
(\frac{1}{2})^{n-1} = \frac{1}{2^5 \cdot 2^4} \\\\
(\frac{1}{2})^{n-1} = \frac{1}{2^{5+4}} \\\\
(\frac{1}{2})^{n-1} = \frac{1}{2^9} \\\\
(\frac{1}{2})^{n-1} = (\frac{1}{2})^9$

$n-1 = 9 \\\\
n = 9+1\\\\
n = 10$

$q \neq 1 \Rightarrow S_n = a_1 \cdot \frac{1-q^n}{1 - q} \\\\
S_{10}= 16 \cdot \frac{1-(\frac{1}{2})^{10}}{1 - \frac{1}{2}} = \frac{16 \cdot (1-\frac{1}{2^{10}})}{\frac{1}{2}} =16 \cdot (1-\frac{1}{2^{10}}) : \frac{1}{2} =\\\\
=(16 - 16 \cdot \frac{1}{2^{10}}) \cdot 2=32 -32 \cdot \frac{1}{2^{10}} =32 -2^5 \cdot \frac{1}{2^{10}} =$

$=32 -\frac{1}{2^{10-5}} =32 -\frac{1}{2^5} =32 -\frac{1}{32} =31\frac{31}{32}$

Suma wynosi: $31\frac{31}{32}$

d)
$\frac{3}{4}+ 1+ \frac{4}{3}+ \frac{16}{9}+ \dots +\frac{1024}{243} \\\\ 
a_1 = \frac{3}{4}, \ a_2 =1 \\\\ 
q = a_2 : a_1 = 1 : \frac{3}{4}= 1\cdot \frac{4}{3} =\frac{4}{3}\\\\ 
a_n = a_1 \cdot q^{n-1} \\\\ 
\frac{3}{4} \cdot (\frac{4}{3})^{n-1} = \frac{1024}{243} \ : \frac{3}{4} \\\\
(\frac{4}{3})^{n-1} = \frac{1024}{243} \cdot \frac{4}{3}$

$(\frac{4}{3})^{n-1} = \frac{4^5}{3^5} \cdot \frac{4}{3} \\\\
(\frac{4}{3})^{n-1} = \frac{4^6}{3^6} \\\\
(\frac{4}{3})^{n-1} = (\frac{4}{3})^6 \\\\
n-1=6 \\\\
n = 6 +1 \\\\
n= 7$

$q \neq 1 \Rightarrow S_n = a_1 \cdot \frac{1-q^n}{1 - q} \\\\
S_7 = \frac{3}{4} \cdot \frac{1-(\frac{4}{3})^7}{1 - \frac{4}{3}} =\frac{3}{4} \cdot \frac{1-(\frac{4}{3})^7}{- \frac{1}{3}} =\frac{3}{4} \cdot (1-\frac{4^7}{3^7}) \cdot (-\frac{3}{1}) =-\frac{9}{4} \cdot (1-\frac{4^7}{3^7})=$

$=-\frac{9}{4}+\frac{4^6}{3^5} = -\frac{9}{4}+\frac{4096}{243} = -\frac{2187}{972}+\frac{4096}{972} = \frac{14197}{972}=14\frac{589}{972}$

Suma wynosi: $14\frac{589}{972}$

1 votes Thanks 1

Zgłoś nadużycie! C)
$16+ 8+ 4+ 2+ ... + \frac{1}{32}$

$a_1=16$

$q=\frac{8}{16}=\frac{1}{2}$

Obliczam $n$
$a_n=a_1q^{n-1}$

$16\cdot (\frac{1}{2})^{n-1}=\frac{1}{32}$

$2^4\cdot (\frac{1}{2})^{n-1}=\frac{1}{2^5}$

$2^4\cdot 2^{-n+1}=2^{-5}$

$2^{4-n+1}=2^{-5}$

$4-n+1=-5$

$-n=-5-4-1$

$-n=-10$

$n=10$

Obliczam $S_{10}$
$S_{10}= \frac{a_1(q^n-1)}{q-1}$

$S_{10}= \frac{16\cdot (( \frac{1}{2} )^{10}-1)}{ \frac{1}{2} -1}$

$S_{10}= \frac{2^4\cdot (^{-10}-1)}{-\frac{1}{2}}$

$S_{10}=-\frac{2^4\cdot (2^{-10}-1)}{2^{-1}}$

$S_{10}=-\frac{2^4\cdot (2^{-10}-1)}{2^{-1}}$

$S_{10}=-2^5\cdot (2^{-10}-1)$

$S_{10}=-32\cdot ( \frac{1}{1024} -1)$

$S_{10}=-32\cdot (-\frac{1023}{1024})$

$S_{10}=\frac{1023}{32}$
=====================
d)
$\frac{3}{4}+1+ \frac{4}{3}+ \frac{16}{9}+...+ \frac{1024}{243}$

$a_1=\frac{3}{4}$

$q= \frac{1}{\frac{3}{4}}=\frac{4}{3}$

Obliczam $n$
$a_n=a_1q^{n-1}$

$\frac{3}{4}\cdot (\frac{4}{3})^{n-1}=\frac{1024}{243}$

$(\frac{4}{3})^{-1}\cdot (\frac{4}{3})^{n-1}=(\frac{4}{243})^5$

$(\frac{4}{3})^{n-1-1}=(\frac{4}{243})^5$

$n-1-1=5$

$n=5+1+1$

$n=7$

Obliczam $S_7$
$S_{n}= \frac{a_1(q^n-1)}{q-1}$

$S_7= \frac{\frac{3}{4}\cdot ((\frac{4}{3})^7-1)}{\frac{4}{3}-1}$

$S_7= \frac{\frac{3}{4}\cdot (\frac{16384}{2187}-1)}{\frac{1}{3}}$

$S_7= \frac{9}{4}\cdot \frac{14197}{2187}$

$S_7= \frac{14197}{972 }$

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