数学 - 無限の世界への一歩 - 数列の極限、無限級数 – 無限級数 - 無限等比級数(漸化式、三角関数)

数学読本〈4〉

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数学読本〈4〉数列の極限，無限級数/順列・組合せ/確率/関数の極限と微分法(松坂 和夫(著)、岩波書店)の第14章(無限の世界への一歩 - 数列の極限、無限級数)、14.3(無限級数)、無限等比級数、問26、27、28、29.を取り組んでみる。

26. 
    $\begin{array}{l}{\displaystyle \sum _{i=1}^m\left(a{r}^{4i-4}-a{r}^{4i-2}\right)}\\ ={\displaystyle \sum _{i=1}^m\left(a{r}^{-4}-a{r}^{-2}\right){\left(r^4\right)}^i}\\ x=\frac{\left(a{r}^{-4}-a{r}^{-2}\right)r^4}{1-r^4}\\ =\frac{a-a{r}^2}{1-r^4}\\ =\frac{a\left(1-r^2\right)}{1-r^4}\\ =\frac{a}{1+r^2}\\ \\ {\displaystyle \sum _{i=1}^m\left(a{r}^{4i-3}-a{r}^{4i-1}\right)}\\ ={\displaystyle \sum _{i=1}^m\left(a{r}^{-3}-a{r}^{-1}\right)}{\left(r^4\right)}^i\\ y=\frac{ar-a{r}^3}{1-r^4}\\ =\frac{ar\left(1-r^2\right)}{1-r^4}\\ =\frac{ar}{1+r^2}\\ \\ \underset{n\to \infty }{\lim }{P}_n=\left(\frac{a}{1+r^2},\frac{ar}{1+r^2}\right)\end{array}$
27. 1. 
       $\begin{array}{l}\frac{1}{2}bc=A_1{C}_1{}^2+\frac{1}{2}{A}_1{C}_1\left(b-A_1{C}_1\right)+\frac{1}{2}{A}_1{C}_1\left(c-A_1{C}_1\right)\\ bc=2A_1{C}_1{}^2+A_1{C}_1\left(b-A_1{C}_1\right)+A_1{C}_1\left(c-A_1{C}_1\right)\\ bc=\left(b+c\right)A_1{C}_1\\ A_1{C}_1=\frac{bc}{b+c}\end{array}$
    2. 
       $\begin{array}{l}{\displaystyle \sum _{k=1}^n{\left({\left(\frac{bc}{b+c}\right)}^k·b\right)}^2}\\ =b^2{\displaystyle \sum _{k=1}^n{\left({\left(\frac{c}{b+c}\right)}^2\right)}^k}\\ \underset{n\to \infty }{\lim }{b}^2{\displaystyle \sum _{k=1}^n{\left({\left(\frac{c}{b+c}\right)}^2\right)}^k}\\ =b^2·{\left(\frac{c}{b+c}\right)}^2·\frac{1}{1-{\left(\frac{c}{b+c}\right)}^2}\\ =b^2·{\left(\frac{c}{b+c}\right)}^2·\frac{{\left(b+c\right)}^2}{{\left(b+c\right)}^2-c^2}\\ =\frac{b^2{c}^2}{b^2+2bc}\\ =\frac{b{c}^2}{b+2c}\end{array}$
28. 
    $\begin{array}{l}{a}_{n+2}-a_{n+1}=-\frac{2}{3}\left(a_{n+1}-a_n\right)\\ b_{n+1}=-\frac{2}{3}{b}_n\\ b_1=a_2-a_1=1-0=1\\ b_n={\left(-\frac{2}{3}\right)}^{n-1}\\ a_n=a_1+{\displaystyle \sum _{k=1}^{n-1}{b}_n}={\displaystyle \sum _{k=1}^{n-1}{\left(-\frac{2}{3}\right)}^{k-1}}\\ \underset{n\to \infty }{\lim }{a}_n=\frac{1}{1+\frac{2}{3}}=\frac{3}{5}\end{array}$
29. 
    $\begin{array}{l}{\displaystyle \sum _{k=0}^m\left({\left(-\frac{1}{2}\right)}^{4k}-{\left(-\frac{1}{2}\right)}^{4k+2}\right)}\\ \frac{1}{1-{\left(-\frac{1}{2}\right)}^4}-\frac{{\left(-\frac{1}{2}\right)}^2}{1-{\left(-\frac{1}{2}\right)}^4}=\frac{1-\frac{1}{4}}{1-\frac{1}{16}}\\ =\frac{3·16}{15·4}\\ =\frac{4}{5}\end{array}$