Infinite Products and Nested Sequences

A question is this type if and only if it involves evaluating infinite products or sequences defined by nested operations like u_{n+1} = f(u_n) with special structure.

3 questions · Hard +2.1

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Edexcel AEA 2022 June Q6
24 marks Hard +2.3
6. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{71cd126f-1c7d-4e37-a26d-7ff98a74fd79-22_481_1139_189_463} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} Figure 2 shows the first few iterations in the construction of a curve, \(L\).
Starting with a straight line \(L _ { 0 }\) of length 4 , the middle half of this line is replaced by three sides of a trapezium above \(L _ { 0 }\) as shown, such that the length of each of these sides is \(\frac { 1 } { 4 }\) of the length of \(L _ { 0 }\) After the first iteration each line segment has length one.
In subsequent iterations, each line segment parallel to \(L _ { 0 }\) similarly has its middle half replaced by three sides of a trapezium above that line segment, with each side \(\frac { 1 } { 4 }\) the length of that line segment. Line segments in \(L _ { n }\) are either parallel to \(L _ { 0 }\) or are sloped.
  1. Show that the length of \(L _ { 2 }\) is \(\frac { 23 } { 4 }\)
  2. Write down the number of
    1. line segments in \(L _ { n }\) that are parallel to \(L _ { 0 }\)
    2. sloped line segments in \(L _ { 2 }\) that are not in \(L _ { 1 }\)
    3. new sloped line segments that are created by the ( \(n + 1\) )th iteration.
  3. Hence find the length of \(L _ { n }\) as \(n \rightarrow \infty\) The area enclosed between \(L _ { 0 }\) and \(L _ { n }\) is \(A _ { n }\)
  4. Find the value of \(A _ { 1 }\)
  5. Find, in terms of \(n\), an expression for \(A _ { n + 1 } - A _ { n }\)
  6. Hence find the value of \(A _ { n }\) as \(n \rightarrow \infty\) The same construction as described above is applied externally to the three sides of an equilateral triangle of side length \(a\).
    Given that the limit of the area of the resulting shape is \(26 \sqrt { 3 }\)
  7. find the value of \(a\).
Edexcel AEA 2006 June Q7
20 marks Hard +2.3
7.
\includegraphics[max width=\textwidth, alt={}, center]{0df09d8a-7478-4679-b117-128ee226db6a-5_648_1590_296_275} The circle \(C _ { 1 }\) has centre \(O\) and radius \(R\). The tangents \(A P\) and \(B P\) to \(C _ { 1 }\) meet at the point \(P\) and angle \(A P B = 2 \alpha , 0 < \alpha < \frac { \pi } { 2 }\). A sequence of circles \(C _ { 1 } , C _ { 2 } , \ldots , C _ { n } , \ldots\) is drawn so that each new circle \(C _ { n + 1 }\) touches each of \(C _ { n } , A P\) and \(B P\) for \(n = 1,2,3 , \ldots\) as shown in Figure 2. The centre of each circle lies on the line \(O P\).
  1. Show that the radii of the circles form a geometric sequence with common ratio $$\frac { 1 - \sin \alpha } { 1 + \sin \alpha }$$
  2. Find, in terms of \(R\) and \(\alpha\), the total area enclosed by all the circles, simplifying your answer. The area inside the quadrilateral \(P A O B\), not enclosed by part of \(C _ { 1 }\) or any of the other circles, is \(S\).
  3. Show that $$S = R ^ { 2 } \left( \alpha + \cot \alpha - \frac { \pi } { 4 } \operatorname { cosec } \alpha - \frac { \pi } { 4 } \sin \alpha \right) .$$
  4. Show that, as \(\alpha\) varies, $$\frac { \mathrm { d } S } { \mathrm {~d} \alpha } = R ^ { 2 } \cot ^ { 2 } \alpha \left( \frac { \pi } { 4 } \cos \alpha - 1 \right)$$
  5. Find, in terms of \(R\), the least value of \(S\) for \(\frac { \pi } { 6 } \leq \alpha \leq \frac { \pi } { 4 }\).
Edexcel AEA 2007 June Q5
15 marks Challenging +1.8
5. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f2290882-b9a4-43ec-a38f-c44d46477242-4_493_1324_279_367} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows part of a sequence \(S _ { 1 } , S _ { 2 } , S _ { 3 } , \ldots\), of model snowflakes. The first term \(S _ { 1 }\) consists of a single square of side \(a\). To obtain \(S _ { 2 }\), the middle third of each edge is replaced with a new square, of side \(\frac { a } { 3 }\), as shown in Figure 1 . Subsequent terms are obtained by replacing the middle third of each external edge of a new square formed in the previous snowflake, by a square \(\frac { 1 } { 3 }\) of the size, as illustrated by \(S _ { 3 }\) in Figure 1.
  1. Deduce that to form \(S _ { 4 } , 36\) new squares of side \(\frac { a } { 27 }\) must be added to \(S _ { 3 }\).
  2. Show that the perimeters of \(S _ { 2 }\) and \(S _ { 3 }\) are \(\frac { 20 a } { 3 }\) and \(\frac { 28 a } { 3 }\) respectively.
  3. Find the perimeter of \(S _ { n }\).
  4. Describe what happens to the perimeter of \(S _ { n }\) as \(n\) increases.
  5. Find the areas of \(S _ { 1 } , S _ { 2 }\) and \(S _ { 3 }\).
  6. Find the smallest value of the constant \(S\) such that the area of \(S _ { n } < S\), for all values of \(n\).
    \includegraphics[max width=\textwidth, alt={}, center]{f2290882-b9a4-43ec-a38f-c44d46477242-5_590_1041_283_588} Figure 2 shows a sketch of the curve \(C\) with equation \(y = \tan \frac { t } { 2 } , \quad 0 \leq t \leq \frac { \pi } { 2 }\).
    The point \(P\) on \(C\) has coordinates \(\left( x , \tan \frac { x } { 2 } \right)\).
    The vertices of rectangle \(R\) are at \(( x , 0 ) , \left( \frac { x } { 2 } , 0 \right) , \left( \frac { x } { 2 } , \tan \frac { x } { 2 } \right)\) and \(\left( x , \tan \frac { x } { 2 } \right)\) as shown in Figure 2.