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1996 Paper 1 Q8
D: 1500.0 B: 1500.0
geometric series recurring decimal fraction conversion sum of geometric series rational numbers proof

  1. By using the formula for the sum of a geometric series, or otherwise, express the number \(0.38383838\ldots\) as a fraction in its lowest terms.
  2. Let \(x\) be a real number which has a recurring decimal expansion \[ x=0\cdot a_{1}a_{2}a_{2}\cdots, \] so that there exists positive integers \(N\) and \(k\) such that \(a_{n+k}=a_{n}\) for all \(n>N.\) Show that \[ x=\frac{b}{10^{N}}+\frac{c}{10^{N}(10^{k}-1)}\,, \] where \(b\) and \(c\) are integers to be found. Deduce that \(x\) is rational.

Solution:

  1. \(\,\) \begin{align*} && 0.383838\ldots &= \frac{3}{10} + \frac{8}{100} + \frac{3}{1000} + \cdots \\ &&&= \frac{38}{100} + \frac{38}{10000} + \cdots \\ &&&= \frac{38}{100} \left (1 + \frac{1}{100} + \frac{1}{100^2} + \cdots \right) \\ &&&= \frac{\frac{38}{100}}{1 - \frac{1}{100}} \\ &&&= \frac{38}{99} \end{align*}
  2. Let \(x = 0\cdot a_{1}a_{2}a_{2}\cdots\) such that there exists \(N\), \(k\) such that \(a_{n+k} = a_n\) for \(n > N\). First notice that \begin{align*} x &= \frac{a_1a_2\cdots a_N}{10\ldots000} + \frac{a_{N+1}}{10^{N+1}} + \cdots \\ &= \frac{b}{10^N} + \frac{a_{N+1}a_{N+2}\cdots a_{N+k}}{10^{N+k}} + \frac{a_{N+1}a_{N+2}\cdots a_{N+2k}}{10^{N+k}} + \cdots \\ &= \frac{b}{10^N} + \frac{c}{10^{N+k}} \left (1 + \frac{1}{10^k} + \cdots \right) \\ &= \frac{b}{10^N} + \frac{c}{10^{N+k}} \frac{1}{1- \frac{1}{10^k}} \\ &= \frac{b}{10^N} + \frac{c}{10^{N+k}} \frac{10^k}{10^k-1} \\ &= \frac{b}{10^N} + \frac{c}{10^N(10^k-1)} \end{align*} where \(b = a_1a_2\cdots a_N\) and \(c = a_{N+1}a_{N+2}\cdots a_{N+k}\). Clearly as the sum of two rational numbers, \(x\) is rational.

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1993 Paper 1 Q2
D: 1500.0 B: 1516.0
geometric series sum of geometric series convergence infinite series pattern recognition partial fractions limiting behaviour

If \(\left|r\right|\neq1,\) show that \[ 1+r^{2}+r^{4}+\cdots+r^{2n}=\frac{1-r^{2n+2}}{1-r^{2}}\,. \] If \(r\neq1,\) find an expression for \(\mathrm{S}_{n}(r),\) where \[ \mathrm{S}_{n}(r)=r+r^{2}+r^{4}+r^{5}+r^{7}+r^{8}+r^{10}+\cdots+r^{3n-1}. \] Show that, if \(\left|r\right|<1,\) then, as \(n\rightarrow\infty,\) \[ \mathrm{S}_{n}(r)\rightarrow\frac{1}{1-r}-\frac{1}{1-r^{3}}\,. \] If \(\left|r\right|\neq1,\) find an expression for \(\mathrm{T}_{n}(r),\) where \[ \mathrm{T}_{n}(r)=1+r^{2}+r^{3}+r^{4}+r^{6}+r^{8}+r^{9}+r^{10}+r^{12}+r^{14}+r^{15}+r^{16}+\cdots+r^{6n}. \] If \(\left|r\right|<1,\) find the limit of \(\mathrm{T}_{n}(r)\) as \(n\rightarrow\infty.\) What happens to \(\mathrm{T}_{n}(r)\) as \(n\rightarrow\infty\) in the three cases \(r>1,r=1\) and \(r=-1\)? In each case give reasons for your answer.

Solution: \begin{align*} && S &= 1 + r^2 + r^4 + \cdots + r^{2n} \\ && r^2S &= \quad \,\,\,\, r^2 + r^4 + \cdots+r^{2n}+r^{2n+2} \\ \Rightarrow && (1-r^2)S &= 1 - r^{2n+2} \\ \Rightarrow && S &= \frac{1-r^{2n+2}}{1-r^2} \end{align*} \begin{align*} && S_n(r) &= r + r^2 + r^4 + r^5 + r^7 + \cdots + r^{3n-1} \\ &&&= 1 + r + r^2 + \cdots + r^{3n} - (1 + r^3 + r^6 + r^{3n}) \\ &&&= \frac{1-r^{3n+1}}{1-r} - \frac{1-r^{3n+3}}{1-r^3} \\ \\ \Rightarrow && \lim_{n \to \infty} S_n(r) &= \frac{1-0}{1-r} - \frac{1-0}{1-r^3} = \frac{1}{1-r} - \frac{1}{1-r^3} \end{align*} \begin{align*} && T_n(r) &= 1 + r^2 + r^3 + r^4 + r^6 + \cdots + r^{6n} \\ &&&= \frac{1-r^{6n+6}}{1-r^6} + \frac{r^2-r^{6n+2}}{1-r^6} + \frac{r^3-r^{6n+3}}{1-r^6} + \frac{r^4-r^{6n+4}}{1-r^6} \\ &&&= \frac{1+r^2+r^3+r^4-r^{6n}(r^2+r^3+r^4+r^6))}{1-r^6} \\ \\ &&\lim_{n \to \infty} T_n(r) &= \frac{1+r^2+r^3+r^4}{1-r^6} \end{align*} If \(r > 1\) clear it diverges. if \(r = 1\) same story. if \(r = -1\) the sums in blocks of \(4\) are all \(1+1-1+1 = 2 > 0\) and so it also diverges.

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