高専の数学解説

第1章 1.29

(1)

数列を計算すると

an→3,2,53,149,⋯a_n \to 3, 2, \frac{5}{3}, \frac{14}{9}, \cdots

bn→−1,−13,−19,⋯∴bn=−(13)n−1b_n \to -1, -\frac{1}{3}, -\frac{1}{9}, \cdots \qquad \therefore b_n = -\left(\frac{1}{3}\right)^{n-1}

よって数列{an}\{a_n\}の一般項は

an=a1+∑k=1n−1bk=3+∑k=1n−1−(13)k−1=3−1−(13)n−11−13=3−1−(13)n−123=32{1+(13)n−1}a_n = a_1 + \sum_{k=1}^{n-1} b_k = 3 + \sum_{k=1}^{n-1} -\left(\frac{1}{3}\right)^{k-1} = 3 - \frac{1-\left(\frac{1}{3}\right)^{n-1}}{1-\frac{1}{3}} = 3 - \frac{1-\left(\frac{1}{3}\right)^{n-1}}{\frac{2}{3}} = \frac{3}{2}\left\{1+\left(\frac{1}{3}\right)^{n-1}\right\}

(2)

3an+1=an+3an+1=13an+1an+1−32=13(an−32)=13bnbn+1=13bn3a_{n+1} = a_n + 3 \qquad a_{n+1} = \frac{1}{3}a_n + 1 \qquad a_{n+1} - \frac{3}{2} = \frac{1}{3}\left(a_n - \frac{3}{2}\right) = \frac{1}{3}b_n \qquad b_{n+1} = \frac{1}{3}b_n

数列bnb_nは、初項b1=3−32=32b_1 = 3 - \frac{3}{2} = \frac{3}{2}、公比13\frac{1}{3}の等比数列だから

bn=32⋅(13)n−1an−32=32⋅(13)n−1∴an=32⋅{1+(13)n−1}b_n = \frac{3}{2}\cdot\left(\frac{1}{3}\right)^{n-1} \qquad a_n - \frac{3}{2} = \frac{3}{2}\cdot\left(\frac{1}{3}\right)^{n-1} \qquad \therefore a_n = \frac{3}{2}\cdot\left\{1+\left(\frac{1}{3}\right)^{n-1}\right\}

解説: r31bn1z