數列與級數/2 等比數列與等比級數-範例

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範例5

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\begin{align} & a_{-10}=25 \\ & a_{0}=25\times r^{10}=30\Rightarrow r^{10}=\frac{6}{5} \\ & a_{20}=30\times \left( r^{10} \right)^{2}=30\times \left( \frac{6}{5} \right)^{2}=30\times 1.44=43.2 \\ \end{align}

範例5類題1

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範例5類題2

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範例5類題3

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範例6

  • 已知：\left\{ \begin{align} & a_{1},a_{2},a_{3},a_{4}\to \text{an arithmetic sequence} \\ & 0<a_{1}<2 \\ & a_{3}=4 \\ & b_{n}=2^{a_{n}} \\ \end{align} \right.
  • 求解：\left\{ \begin{align} & b_{1},b_{2},b_{3},b_{4}\to \text{a geometric sequence} \\ & b_{1}<b_{2} \\ & b_{2}>4 \\ & b_{4}>32 \\ & b_{2}\times b_{4}=256 \\ \end{align} \right.

答案

詳解

  1. 已知：\left\{ \begin{align} & a_{1},a_{2},a_{3},a_{4}\to \text{an arithmetic sequence} \\ & 0<a_{1}<2 \\ & a_{3}=4 \\ & b_{n}=2^{a_{n}} \\ \end{align} \right.
  2. a_{3}=4=a_{1}+2d\Rightarrow d=\frac{1}{2}\left( 4-a_{1} \right)
  3. 0<a_{1}<2\Rightarrow 0>-a_{1}>-2\Rightarrow 4>4-a_{1}>2\Rightarrow 2>\frac{1}{2}\left( 4-a_{1} \right)>1\Rightarrow 2>d>1
  4. b_{n}=2^{a_{n}}\Rightarrow \frac{b_{k+1}}{b_{k}}=\frac{2^{a_{k+1}}}{2^{a_{k}}}=2^{a_{k+1}-a_{k}}=2^{d},\forall k\in N
    \Rightarrow b_{1},b_{2},b_{3},b_{4}\to \text{a geometric sequence with common ratio }r=2^{d}\text{,}2>d>1\text{ }\!\!\#\!\!\text{ }
    \Rightarrow \left\{ \begin{align} & 2>d>1\Rightarrow 2^{2}>2^{d}>2^{1}\Rightarrow 4>2^{d}>2 \\ & b_{2}=b_{1}\times 2^{d}\Rightarrow b_{2}>b_{1}\# \\ \end{align} \right.
  5. \left\{ \begin{align} & a_{3}=4 \\ & b_{n}=2^{a_{n}} \\ \end{align} \right.\Rightarrow b_{3}=2^{a_{3}}=2^{4}=16
  6. \left\{ \begin{align} & b_{2}=\frac{b_{3}}{r}=\frac{16}{2^{d}} \\ & 4>2^{d}>2\Rightarrow \frac{16}{4}<\frac{16}{2^{d}}<\frac{16}{2} \\ \end{align} \right.\Rightarrow 4<b_{2}<8\#
  7. \left\{ \begin{align} & \left\{ \begin{align} & b_{4}=b_{3}\times 2^{d} \\ & b_{3}=16 \\ \end{align} \right.\Rightarrow b_{4}=16\times 2^{d} \\ & 4>2^{d}>2\Rightarrow 16\times 4>16\times 2^{d}>16\times 2 \\ \end{align} \right.\Rightarrow 64>b_{4}>32\#
  8. b_{k-1}\times b_{k+1}=b_{k}^{2}\Rightarrow b_{2}\times b_{4}=b_{3}^{2}=16^{2}=256\# ，提示：等比中項定理

範例6類題1

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範例6類題2

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範例6類題3

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範例7

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\begin{align} & (1) \\ & S_{n}=\frac{a(1-r^{n})}{1-r}=48 \\ & S_{2n}=\frac{a(1-r^{2n})}{1-r}=60 \\ & \frac{S_{n}}{S_{2n}}=\frac{(1-r^{n})}{(1-r^{2n})}=\frac{(1-r^{n})}{(1-r^{2n})}=\frac{(1-r^{n})}{(1-r^{n})(1+r^{n})}=\frac{1}{1+r^{n}}=\frac{48}{60} \\ & \therefore \frac{1}{1+r^{n}}=\frac{4}{5}\Rightarrow r^{n}=\frac{1}{4} \\ & (2) \\ & S_{n}=\frac{a(1-\frac{1}{4})}{1-r}=48\therefore \frac{a}{1-r}=64 \\ & (3) \\ & \frac{a\left( 1-r^{3n} \right)}{1-r}=64\times \left[ 1-\left( \frac{1}{4} \right)^{3} \right]=63 \\ \end{align}

範例7類題

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