جواب تمرین صفحه 83 درس 3 ریاضی و آمار دوازدهم انسانی (الگوهای غیر خطی)

 1 \({a_{n + 1}} = {\left( {{a_n}} \right)^2}\;\;\;\;{a_1} = \frac{1}{2}\)

\( \Rightarrow \;\;\:\frac{1}{2}\;\;\:,\;\;\:\frac{1}{4}\;\;\:,\;\;\:\frac{1}{{16}}\;\;\:,\;\;\: \cdots \)

2 \({a_{n + 1}} = \frac{2}{3}{a_n}\;\;\;\;{a_1} = \frac{1}{2}\)

\( \Rightarrow \;\;\:\frac{1}{2}\;\;\:,\;\;\:\frac{1}{3}\;\;\:,\;\;\:\frac{2}{9}\;\;\:,\;\;\: \cdots \)

دنباله هندسی

3 \({a_{n + 1}} = \frac{1}{{1 + {a_n}}}\;\;\;\;{a_1} = 1\)

\( \Rightarrow \;\;\:1\;\;\:,\;\;\:\frac{1}{2}\;\;\:,\;\;\:\frac{2}{3}\;\;\:,\;\;\: \cdots \)

4 \({a_{n + 1}} = 2{a_n}\;\;\;\;{a_1} = 1\)

\( \Rightarrow \;\;\:1\;\;\:,\;\;\:2\;\;\:,\;\;\:4\;\;\:,\;\;\: \cdots \)

دنباله هندسی

دنباله \(a\;,\;b\;,\;c\)  ، یک دنباله هندسی است؛ بنابراین :

\(\begin{array}{l}a\;,\;b\;,\;c\;\; \Rightarrow \;\;r = \frac{b}{a} = \frac{c}{b}\\\\ \Rightarrow \;\;{b^2} = ac\end{array}\)

\(\begin{array}{l}x\:,\:x + 2\:,\:x + 3\;\;\: \Rightarrow \;{(x + 2)^2} = x\:(x + 3)\\\\ \Rightarrow {x^2} + 4x + 4 = {x^2} + 3x \Rightarrow x + 4 = 0\\\\ \Rightarrow x = - 4\end{array}\)

الف

\(\begin{array}{l}{a_n} = {10^{12}}{(\frac{4}{{10}})^{n - 1}}\quad \\\\ \Rightarrow {a_3} = {10^{12}}{(\frac{4}{{10}})^{3 - 1}} = {10^{12}}{(\frac{4}{{10}})^2}\\\\ = 1/6 \times {10^{11}}\end{array}\)

ب

\(\begin{array}{l}{10^{12}}{(\frac{4}{{10}})^{n - 1}} < 7 \times {10^6} \Rightarrow {(\frac{4}{{10}})^{n - 1}} < 7 \times {10^{ - 6}}\\\\n = 2 \Rightarrow 0/4 > 7 \times {10^{ - 6}}\\\\n = 3 \Rightarrow {(0/4)^2} = 0/16 > 7 \times {10^{ - 6}}\\\quad \vdots \\n = 13 \Rightarrow {(0/4)^{12}} = 1/6777 \times {10^{ - 5}} > 7 \times {10^{ - 6}}\\\\n = 14 \Rightarrow {(0/4)^{13}} = 6/7108 \times {10^{ - 6}} < 7 \times {10^{ - 6}}\\\\ \Rightarrow \quad n = 14\end{array}\)

\(\begin{array}{l}a\;,\mathop {\underline {\quad \cdots \quad } }\limits_{n\;Sentence} ,\;b\quad \Rightarrow \quad n + 2\;Sentence\quad \\\\ \Rightarrow \quad \left\{ \begin{array}{l}{a_{n + 2}} = a\;{r^{n + 1}}\\{a_{n + 2}} = b\end{array} \right. \Rightarrow a\;{r^{n + 1}} = b\\\\ \Rightarrow {r^{n + 1}} = \frac{b}{a}\end{array}\)

sentecnce به معنی جمله است.

\(\begin{array}{l}{a_1}\;,\;{a_2}\;,\;\mathop {\underline{\underline {{a_3}}} }\limits_{27} \;,\;{a_4}\;,\;\mathop {\underline{\underline {{a_5}}} }\limits_{243} \;,\; \cdots \quad \\\\ \Rightarrow \quad \frac{{{a_5}}}{{{a_3}}} = \frac{{{a_1}\;{r^4}}}{{{a_1}\;{r^2}}} = {r^2} = \frac{{243}}{{27}} = 9\\\;\;\quad \;\\ \Rightarrow {a_7} = {a_1}\;{r^6} = {a_1}\;{r^4} \times {r^2} = 243 \times 9 = 2187\end{array}\)

\({x^2} = a\;b\)

x واسطه هندسی بین a و b است؛ پس دنباله هندسی آن \(a\;,\;x\;,\;b\) یا \(b\;,\;x\;,\;a\)  است. در نتیجه جواب گزینه ب می باشد.

\(\begin{array}{l}{a_n} = 15000\;{(\frac{{85}}{{100}})^{n - 1}}\\\\ \Rightarrow {a_5} = 15000\;{(\frac{{85}}{{100}})^4} = 7830/09375\end{array}\)

الف

\(\begin{array}{l}2\;,\;4\;,\;8\;,\; \cdots \quad \Rightarrow \quad {a_1} = 2\quad ,\quad r = 2\\\\{a_n} = 2 \times {2^{n - 1}} = {2^n}\quad \Rightarrow \quad {a_{10}} = {2^{10}} = 1024\end{array}\)

ب

\({S_n} = 2 \times \frac{{1 - {2^{10}}}}{{1 - 2}} = - 2 + {2^{11}} = - 2 + 2048 = 2046\)

الف \(1 + 4 + 16 + … + 4096\)

\(\begin{array}{*{20}{l}}\begin{array}{l}{a_1} = 1\;\;\:,\;\;\:r = 4\;\;\: \Rightarrow \;\;\:{a_n} = {4^{n - 1}} = 4096 = {4^6}\\\\ \Rightarrow n - 1 = 6 \Rightarrow n = 7\end{array}\\\begin{array}{l}\\{S_7} = \frac{{1\:(1 - {4^7})}}{{1 - 4}} = \frac{{ - 16383}}{{ - 3}} = 5461\end{array}\end{array}\)

ب \(\frac{1}{5} + \frac{1}{{10}} + \frac{1}{{20}} + \;...\; + \frac{1}{{640}}\)

\(\begin{array}{*{20}{l}}\begin{array}{l}{a_1} = \frac{1}{5}\;\;\:,\;\;\:r = \frac{1}{2}\;\; \Rightarrow \;\:{a_n} = \frac{1}{5}{(\frac{1}{2})^{n - 1}} = \frac{1}{{640}}\\\\ \Rightarrow {(\frac{1}{2})^{n - 1}} = \frac{1}{{128}} = {(\frac{1}{2})^7} \Rightarrow n - 1 = 7 \Rightarrow n = 8\end{array}\\\begin{array}{l}\\{S_8} = \frac{{\frac{1}{5}\:(1 - {{(\frac{1}{2})}^8})}}{{1 - \frac{1}{2}}} = \frac{{2 - {{(\frac{1}{2})}^7}}}{5} = \frac{{2 - \frac{1}{{128}}}}{5} = \frac{{255}}{{640}} = \frac{{51}}{{128}}\end{array}\end{array}\)

\(\begin{array}{*{20}{l}}{{a_1} = 1536\;\;\:,\;\;\:r = \frac{1}{2}\;\;\:,\;\;\:{a_n} = 6\;\;\:,\;\;\: \Rightarrow n = ?}\\\begin{array}{l}\\{a_n} = 1536\:{(\frac{1}{2})^{n - 1}} = 6 \Rightarrow {(\frac{1}{2})^{n - 1}} = \frac{1}{{256}} = {(\frac{1}{2})^8}\\\\ \Rightarrow n - 1 = 8 \Rightarrow n = 9\end{array}\\\begin{array}{l}\\{S_9} = \frac{{1536\:(1 - {{(\frac{1}{2})}^9})}}{{1 - \frac{1}{2}}} = \frac{{1536\: - \frac{{1536}}{{512}}}}{{\frac{1}{2}}}\\\\ = 2 \times 1536 - \frac{{1536}}{{256}} = 3072 - 6 = 3066\end{array}\end{array}\)

الف زیرا مساحت هر قسمت از ضرب مساحت قسمت قبلی در \(\frac{1}{2}\) به دست می آید.

ب 

\({S_n} = \frac{1}{4} \times \frac{{1 - {{(\frac{1}{4})}^n}}}{{1 - \frac{1}{4}}} = \frac{{1 - {{(\frac{1}{4})}^n}}}{3}\)

پ 

\({S_6} = \frac{{1 - {{(\frac{1}{4})}^6}}}{3} = \frac{{1 - \frac{1}{{4096}}}}{3} = \frac{{4095}}{{12288}}\)