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AI HL · 12.1

Sequences & Series

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12.1.9

Infinite series in context: bouncing balls and nested squares

12.1.9

Infinite series in context: bouncing balls and nested squares

Target: I can write the height of a bouncing ball after n bounces as a term of a geometric sequence. (+5 more)

By the end of this section

Key points

Setting up a bouncing ball

  • A ball dropped from height h that rebounds to a fraction r of its last height reaches \(hr\), \(hr^2\), \(hr^3\), \(\dots\) The height after the nth bounce is \(u_n=hr^n\). The drop itself is the starting height, not a bounce.
  • Check what the first term is and where counting starts. After 6 bounces the power is 6, not 5.
  • For time, each bounce is a rise and a fall of equal time. The first fall happens only once. A drop time of t with rebound factor r gives \(t+2tr+2tr^2+\dots\)
  • In a "show that" question, work out the first two or three flights from the words of the question, then say what the pattern is.

Finite sums

  • \(S_n=\frac{u_1(1-r^n)}{1-r}\) for a geometric series with n terms.
  • If the first term does not fit the pattern, as with the single fall, keep it apart. Add it to the sum of the other \(n-1\) terms, which are geometric.
  • Check your formula by putting in \(n=1\) and \(n=2\) and comparing with the terms.

Nested squares

  • Joining the midpoints of the sides of a square makes a new square of half the area. The ratio of areas is \(\frac12\), not the ratio of the sides.
  • Lengths scale by k, areas by \(k^2\). Work with areas, as it avoids square roots.
  • Write down the shaded area of stage 1, then stage 2. The ratio between them is r. Then use \(S_n\) for n stages.

Coming to rest: sum to infinity

  • A ball bounces infinitely often yet is at rest after a finite time and distance. This works because \(|r|<1\), so the series has a sum.
  • Total time: the first fall plus \(S_\infty\) of the remaining flights, or write the whole series and take out the first term.
  • Total distance: the drop plus twice the sum of the bounce heights, because each bounce is up and then down. If the first bounce height is \(u_1\), the distance is \(h+\frac{2u_1}{1-r}\).

Shaded area to infinity, and unknown x

  • If the nested squares go on for ever, the shaded area is \(S_\infty=\frac{u_1}{1-r}\) with the same \(u_1\) and r as for the finite sum. Compare with the area of the first square as a check.
  • To find x when an infinite series has a given sum, write the ratio in terms of x, use \(\frac{u_1}{1-r}\) with the first term in terms of x, set it equal to the sum, and solve.
  • Check that the x you find gives \(|r|<1\).

On the calculator

Checking a sum with a table

  • Enter the formula for the nth term in a table and read the terms for \(n=1,2,3,\dots\)
  • Add a running total column, or use the sum command, to compare with your formula for \(S_n\).

Screenshot and video to come

Solving \(S_n\) against a value

  • Enter \(S_n\) as a function of n.
  • Use a table or the solver to find where it passes the given value.
  • n must be a whole number: choose the first whole number that works and check it in the formula.

Screenshot and video to come

Explore: the bouncing ball

Drag the orange ball up or down to set the drop height, and move the slider to set how high it rebounds. Work out the total distance first, then press Show answer.

The first six bounces. Heights and distances are to 3 significant figures.
BounceHeight (m)Distance so far (m)

The total is worked out from the unrounded values.

Common ratio, r
First bounce, u₁ = hr
Check: h + 2u₁ / (1 − r)

Drag the ball or move the slider.

  1. The ball bounces for ever. Why does it still travel a finite distance?
  2. Set the rebound to 50%. The falls add up to 2h. Why do the rises add up to h?
  3. Double the drop height. What happens to the total distance? Why?
  4. Move the rebound towards 95%. What happens to the total, and why does the series need r < 1?

Explore: the nested squares

Join the midpoints of the sides of the square to make a new square. The four corners left over are shaded. Press Next stage, and before each press say how much area will be shaded. Then press Show answer to check.

Orange: the corners shaded at this stage. Teal: the corners shaded before. Yellow: the middle square, not yet shaded.

The shaded area at each stage, in cm².
StageShadedTotal shaded

Press Next stage.

  1. How does the area of each new square compare with the one before it? Why is it not the sides that halve?
  2. What is the common ratio of the shaded areas? Is it the same at every stage?
  3. The total shaded keeps growing. What number does it never pass, and why?
  4. Change the side of the first square. What changes in the table, and what stays the same?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Showing the series for the time

A ball is dropped and takes 2 s to hit the ground. Each time it bounces it takes 75% of the time of the previous fall to rise, and the same time to fall back. Show that the total time of the motion is given by \(2+4(0.75)+4(0.75)^2+\dots\) seconds.

Working
  1. First fall

    \(2\) s

    Counted once.

  2. First bounce

    rise \(0.75\times 2=1.5\), fall \(1.5\): total \(3=4(0.75)\)

    A rise and a fall of equal time.

  3. Second bounce

    rise \(0.75\times1.5=1.125\), fall \(1.125\): total \(2.25=4(0.75)^2\)

    Each flight is 0.75 of the one before.

  4. The pattern

    \(2+4(0.75)+4(0.75)^2+\dots\)

    After the first fall, each bounce gives \(4(0.75)^k\) seconds.

Video walkthrough coming
A formula for \(S_n\)

For the series \(2+4(0.75)+4(0.75)^2+\dots\) find a formula for \(S_n\), and use it to find \(S_5\).

Working
  1. Split off the first term

    \(S_n=2+\left[3+3(0.75)+\dots\right]\)

    The terms after the 2 are \(4(0.75)=3\), then \(3(0.75)\), and so on. There are \(n-1\) of them.

  2. Geometric sum

    \(S_n=2+\frac{3(1-0.75^{n-1})}{1-0.75}\)

    \(u_1=3,\ r=0.75\), and \(n-1\) terms.

  3. Tidy

    \(S_n=14-12(0.75)^{n-1}\)

    \(2+12=14\). Check \(n=1\): \(14-12=2\).

  4. \(S_5\)

    \(S_5=14-12(0.75)^4=10.2\) s

    From the calculator, 3 significant figures.

Video walkthrough coming
Time until the ball is at rest

For the ball in the last two examples, find the total time until it comes to rest.

Working
  1. Split the series

    \(2+\left[4(0.75)+4(0.75)^2+\dots\right]\)

    The first fall is not part of the geometric series.

  2. Geometric part

    \(u_1=3,\ r=0.75\)

    \(|r|<1\), so the sum exists.

  3. Sum to infinity

    \(S_\infty=\frac{3}{1-0.75}=12\)

    Time for all the bounces.

  4. Add the first fall

    \(2+12=14\) s

    Same as the limit of \(14-12(0.75)^{n-1}\).

Video walkthrough coming
Height after a bounce

A ball is dropped from a height of 12 m. After each bounce it rises to 65% of the height of the bounce before. Find the height it reaches after the 6th bounce.

Working
  1. The heights

    \(12\times0.65,\ 12\times0.65^2,\ \dots\)

    The first bounce reaches \(7.8\) m.

  2. nth term

    \(u_n=12(0.65)^n\)

    Geometric, with \(r=0.65\). The power is the number of bounces.

  3. Sixth bounce

    \(u_6=12(0.65)^6=0.905\) m

    From the calculator, 3 significant figures.

Video walkthrough coming
Total distance until the ball stops

For the same ball, dropped from 12 m with each bounce 65% of the one before, find the total distance it travels before it comes to rest.

Working
  1. The bounce heights

    \(u_1=12\times0.65=7.8,\quad r=0.65\)

    These are the heights reached, going up.

  2. Up and down

    \(\frac{2u_1}{1-r}=\frac{2\times7.8}{0.35}=44.57\dots\)

    Each height is travelled twice.

  3. Add the drop

    \(12+44.57\dots=56.6\) m

    The drop is travelled once.

Video walkthrough coming
Shaded area of nested squares

A square has sides of 12 cm. The midpoints of its sides are joined to make a new square. The four corner triangles that are left over are shaded. The same is done in the new square, and again, as shown. Find the total shaded area after 5 stages.

Working
  1. Stage 1

    Square \(12^2=144\). Inner square has half the area: 72. Shaded corners \(144-72=72\)

    The four triangles make up the other half.

  2. Stage 2

    Inner square 72. Next square 36. Shaded corners \(72-36=36\)

    Each stage is half the one before, so \(r=\frac12\).

  3. Geometric series

    \(72+36+18+\dots\): \(u_1=72,\ r=\frac12\)

    The ratio of areas is \(\frac12\), not the ratio of sides.

  4. Five stages

    \(S_5=\frac{72\left(1-\left(\frac12\right)^5\right)}{1-\frac12}=139.5\)

    Total shaded area 139.5 cm².

Video walkthrough coming
Shaded area for ever

The process in the last example goes on for ever. Find the total shaded area.

Working
  1. Same series

    \(72+36+18+\dots\), \(u_1=72,\ r=\frac12\)

    \(|r|<1\), so the sum exists.

  2. Sum to infinity

    \(S_\infty=\frac{72}{1-\frac12}=144\)

    Total shaded area 144 cm².

  3. Check

    The original square has area \(12^2=144\)

    The triangles fill the whole square in the limit, so 144 is sensible.

Video walkthrough coming
Finding x

Given that \(x+\frac{x^2}{3}+\frac{x^3}{9}+\dots=6\), find the value of x.

Working
  1. First term and ratio

    \(u_1=x,\quad r=\frac{x^2/3}{x}=\frac x3\)

    Second term divided by first.

  2. Write the sum

    \(\frac{x}{1-\frac x3}=6\)

    Use \(S_\infty=\frac{u_1}{1-r}\).

  3. Solve

    \(x=6-2x\Rightarrow 3x=6\Rightarrow x=2\)

    Multiply by \(1-\frac x3\), or use the calculator's solve.

  4. Check

    \(r=\frac23\), so \(|r|<1\)

    The sum exists for this x.

Video walkthrough coming