Why bother?

Some­times you do not add the same amount each time; you mul­ti­ply by the same amount. Money in a bank at com­pound inter­est behaves like this, and so does a ball that keeps bounc­ing a lit­tle lower. Such sequences are geo­met­ric pro­gres­sions. We will find the gen­eral term and the sum of nn terms, see why an infi­nite sum makes sense when ∣r∣<1\lvert r \rvert < 1, meet the geo­met­ric mean and com­pare it with the arith­metic mean, and end with the sums of squares and cubes.

Geo­met­ric pro­gres­sions

In a GP each term is the pre­vi­ous one times a con­stant com­mon ratio rr: a,ar,ar2,…a, ar, ar^2, \ldots

an=arn−1,Sn=a(rn−1)r−1=a(1−rn)1−r(r≠1),\displaystyle a_n = ar^{n-1}, \qquad S_n = \frac{a(r^n - 1)}{r - 1} = \frac{a(1 - r^n)}{1 - r} \quad (r \ne 1),

and Sn=naS_n = na if r=1r = 1. For the proof, write the sum and r times the sum one below the other and sub­tract; almost every­thing can­cels and you are left with Sn−rSn=a−arnS_n - rS_n = a - ar^n.

Exam­ple 1. In 5,15,45,…5, 15, 45, \ldots: a8=5⋅37=10,935a_8 = 5 \cdot 3^7 = 10{,}935; S8=5(38−1)2=16,400\displaystyle S_8 = \tfrac{5(3^8 - 1)}{2} = 16{,}400.

Exam­ple 2. Which term of 14,12,1,…\displaystyle \tfrac{1}{4}, \tfrac{1}{2}, 1, \ldots is 128128? Set up the equa­tion 14⋅2n−1=128\displaystyle \tfrac{1}{4} \cdot 2^{n-1} = 128, 2n−1=5122^{n-1} = 512, n=10n = 10.

Exam­ple 3. The 33rd term is 1212 and the 66th is 9696. Divid­ing one by the other, r3=8r^3 = 8, r=2r = 2, a=3a = 3.

Exam­ple 4. 0.4+0.44+0.444+⋯0.4 + 0.44 + 0.444 + \cdots to nn terms =49[(1−0.1)+(1−0.01)+⋯ ]=49[n−19(1−10−n)]\displaystyle = \tfrac{4}{9}[(1 - 0.1) + (1 - 0.01) + \cdots] = \tfrac{4}{9}\left[n - \tfrac{1}{9}(1 - 10^{-n})\right].

Exam­ple 5. ₹20,00020{,}000 kept at 8%8\% com­pound inter­est a year grows to 20,000(1.08)n20{,}000(1.08)^n after nn years. That is just a GP with ratio 1.081.08.

Adding for­ever: the infi­nite GP

Can you add infi­nitely many num­bers and get a finite answer? Yes, when ∣r∣<1\lvert r \rvert < 1. Then rn→0r^n \to 0, and

S∞=a1−r.\displaystyle S_\infty = \frac{a}{1 - r}.

Exam­ple 6. 6−2+23−⋯=61+1/3=92\displaystyle 6 - 2 + \tfrac{2}{3} - \cdots = \tfrac{6}{1 + 1/3} = \tfrac{9}{2}. 0.27‾=27100⋅11−1/100=2799=311\displaystyle 0.\overline{27} = \tfrac{27}{100} \cdot \tfrac{1}{1 - 1/100} = \tfrac{27}{99} = \tfrac{3}{11}.

Plot of the partial sums of 6 - 2 + 2/3 - ... for 1 to 10 terms: 6, 4, 4.67, 4.44 and so on, swinging above and below a dashed line at 9/2 and closing in on it.
Exam­ple 6: the par­tial sums set­tle on 9/2

Exam­ple 7. A ball dropped from 1212 m rebounds to 23\displaystyle \tfrac{2}{3} of its height each time. Remem­ber it goes up and comes down on every bounce, so the total dis­tance is 12+2×81−2/3=12+48=60\displaystyle 12 + 2 \times \tfrac{8}{1 - 2/3} = 12 + 48 = 60 m.

A ball dropped from 12 m bounces in ever smaller arcs; the first rebound heights are marked 8 m, 16/3 m and 32/9 m, each two-thirds of the one before.
Exam­ple 7: each rebound is two-thirds of the pre­vi­ous height

Geo­met­ric mean

The GM of pos­i­tive aa and bb is G=abG = \sqrt{ab}, and then a,G,ba, G, b is a GP. To slip nn num­bers between aa and bb mak­ing a GP, take r=(ba)1/(n+1)\displaystyle r = \left(\tfrac{b}{a}\right)^{1/(n+1)}.

AM ≥\ge GM. A−G=a+b2−ab=(a−b)22≥0\displaystyle A - G = \tfrac{a + b}{2} - \sqrt{ab} = \tfrac{(\sqrt{a} - \sqrt{b})^2}{2} \ge 0, with equal­ity only when a=ba = b. Keep this one in your pocket; it turns up in many inequal­ity prob­lems.

Exam­ple 8. Two num­bers have AM 1010 and GM 88. That tells us a+b=20a + b = 20, ab=64ab = 64, so the num­bers are 1616 and 44.

Sums of pow­ers worth remem­ber­ing

∑k=1nk=n(n+1)2,∑k=1nk2=n(n+1)(2n+1)6,∑k=1nk3=[n(n+1)2]2.\displaystyle \sum_{k=1}^n k = \frac{n(n + 1)}{2}, \quad \sum_{k=1}^n k^2 = \frac{n(n + 1)(2n + 1)}{6}, \quad \sum_{k=1}^n k^3 = \left[\frac{n(n + 1)}{2}\right]^2.

Exam­ple 9. ∑k=1nk(k+2)=∑k2+2∑k=n(n+1)(2n+7)6\displaystyle \sum_{k=1}^{n} k(k + 2) = \sum k^2 + 2\sum k = \tfrac{n(n + 1)(2n + 7)}{6}.

Try these your­self

  1. Find the 99th term and the sum of 99 terms of 2,−6,18,…2, -6, 18, \ldots
  2. Which term of 2,2,22,…\sqrt{2}, 2, 2\sqrt{2}, \ldots is 6464?
  3. The sum of the first three terms of a GP is 1312\displaystyle \tfrac{13}{12} and their prod­uct is −1-1. Find the terms.
  4. Find the sum to infin­ity: 1−13+19−⋯\displaystyle 1 - \tfrac{1}{3} + \tfrac{1}{9} - \cdots; 0.6‾0.\overline{6} as a frac­tion.
  5. Find nn so that 3+6+12+⋯3 + 6 + 12 + \cdots to nn terms equals 765765.
  6. Insert 33 geo­met­ric means between 22 and 162162.
  7. The AM and GM of two pos­i­tive num­bers are 2525 and 1515. Find the num­bers.
  8. Eval­u­ate ∑k=112(3k2−k)\displaystyle \sum_{k=1}^{12} (3k^2 - k) and 13+23+⋯+1031^3 + 2^3 + \cdots + 10^3.
  9. If a,b,ca, b, c are in GP, show that a2+b2,ab+bc,b2+c2a^2 + b^2, ab + bc, b^2 + c^2 are also in GP.
  10. A cul­ture of bac­te­ria dou­bles every hour. Start­ing with 5050, how many after 88 hours?

Answers to check against

Show answers
  1. 2⋅(−3)8=13,1222 \cdot (-3)^8 = 13{,}122; 2(1−(−3)9)4=9,842\displaystyle \tfrac{2(1 - (-3)^9)}{4} = 9{,}842.
  2. 2n=64=212\sqrt{2}^n = 64 = \sqrt{2}^{12}: 1212th.
  3. Mid­dle term bb with b3=−1b^3 = -1, b=−1b = -1; −1r−1−r=1312\displaystyle \tfrac{-1}{r} - 1 - r = \tfrac{13}{12} gives 12r2+25r+12=012r^2 + 25r + 12 = 0, r=−34\displaystyle r = -\tfrac{3}{4} or −43\displaystyle -\tfrac{4}{3}: terms 43,−1,34\displaystyle \tfrac{4}{3}, -1, \tfrac{3}{4} (either order).
  4. 34\displaystyle \tfrac{3}{4}; 23\displaystyle \tfrac{2}{3}.
  5. 3(2n−1)=7653(2^n - 1) = 765, 2n=2562^n = 256, n=8n = 8.
  6. r4=81r^4 = 81, r=3r = 3: 6,18,546, 18, 54.
  7. a+b=50a + b = 50, ab=225ab = 225: 4545 and 55.
  8. 3⋅650−78=18723 \cdot 650 - 78 = 1872; 30253025.
  9. With b2=acb^2 = ac: (ab+bc)2=b2(a+c)2=ac(a+c)2(ab + bc)^2 = b^2(a + c)^2 = ac(a + c)^2 and (a2+b2)(b2+c2)=(a2+ac)(ac+c2)=ac(a+c)2(a^2 + b^2)(b^2 + c^2) = (a^2 + ac)(ac + c^2) = ac(a + c)^2.
  10. 50×28=12,80050 \times 2^8 = 12{,}800.