Part I. Section III. Chapter 11. “Of Geometrical Progressions.”

505 A series of numbers, which are always becoming a certain number of times greater, or less, is called a geometrical progression, because each term is constantly to the following one in the same geometrical ratio: and the number which expresses how many times each term is greater than the preceding, is called the exponent, or ratio. Thus, when the first term is 1 and the exponent, or ratio, is 2, the geometrical progression becomes,

Terms 1 2 3 4 5 6 7 8 9 etc.
Progression 1, 2, 4, 8, 16, 32, 64, 128, 256, etc.

The numbers 1, 2, 3, etc. always marking the place which each term holds in the progression.

506 If we suppose, in general, the first term to be \(a\), and the ratio \(b\), we have the following geometrical progression

Terms 1 2 3 4 5 6 7 8 ⋯ \(n\)
Progression \(a\), \(ab\), \(ab^2\), \(ab^3\), \(ab^4\), \(ab^5\), \(ab^6\), \(ab^7\), ⋯ \(ab^{n-1}\).

So that, when this progression consists of \(n\) terms, the last term is \(ab^{n-1}\). We must, however, remark here, that if the ratio \(b\) be greater than unity, the terms increase continually; if \(b=1\), the terms are all equal; lastly, if \(b\) be less than 1, or a fraction, the terms continually decrease. Thus, when \(a=1\), and \(b=\frac{1}{2}\), we have this geometrical progression:

1, ½, ¼, ⅛, ¹⁄₁₆, ¹⁄₃₂, ¹⁄₆₄, ¹⁄₁₂₈, etc.

507 Here therefore we have to consider:

  1. The first term, which we have called \(a\).
  2. The exponent, which we call \(b\).
  3. The number of terms, which we have expressed by \(n\).
  4. And the last term, which, we have already seen, is expressed by \(ab^{n-1}\).

So that, when the first three of these are given, the last term is found by multiplying the \(n-1\) power of \(b\), or \(b^{n-1}\), by the first term \(a\).

If, therefore, the 50th term of the geometrical progression 1, 2, 4, 8, etc. were required, we should have \(a=1\), \(b = 2\), and \(n=50\); consequently the 50th term would be 2⁴⁹; and as 2⁹=512, we shall have 2¹⁰=1024; wherefore the square of 2¹⁰, or 2²⁰, =1048576, and the square of this number, which is 1099511627776=2⁴⁰. Multiplying therefore this value of 2⁴⁰ by 2⁹, or 512, we have

2⁴⁹=562949953421312

for the 50th term.

508 One of the principal questions which occurs on this subject, is to find the sum of all the terms of a geometrical progression; we shall therefore explain the method of doing this. Let there be given, first, the following progression, consisting of ten terms:

1, 2, 4, 8, 16, 32, 64, 128, 256, 512,

the sum of which we shall represent by \(s\), so that

\[s=1+2+4+8+16+32+64+128+256+512;\]

doubling both sides, we shall have

\[2s=2+4+8+16+32+64+128+256+512+1024;\]

and subtracting from this the progression represented by \(s\), there remains \(s=1024-1=1023\); wherefore the sum required is 1023.

509 Suppose now, in the same progression, that the number of terms is undetermined, that is, let them be generally represented by \(n\), so that the sum in question, or

\[s=1+2+2^2+2^3+2^4+\cdots+2^{n-1}.\]

If we multiply by 2, we have

\[2s=2+2^2+2^3+2^4+\cdots+2^n,\]

then subtracting from this equation the preceding one, we have \(s=2^n-1-1\). It is evident, therefore, that the sum required is found, by multiplying the last term, \(2^{n-1}\), by the exponent 2, in order to have \(2^n\), and subtracting unity from that product.

510 This is made still more evident by the following examples, in which we substitute successively for \(n\), the numbers 1, 2, 3, 4, etc.

1 = 1;
1 + 2 = 3;
1 + 2 + 4 = 7;
1 + 2 + 4 + 8 = 15;
1 + 2 + 4 + 8 + 16 = 31;
1 + 2 + 4 + 8 + 16 + 32 = 63, etc.

511 On this subject, the following question is generally proposed. A man offers to sell his horse on the following condition; that is, he demands 1 penny for the first nail, 2 for the second, 4 for the third, 8 for the fourth, and so on, doubling the price of each succeeding nail. It is required to find the price of the horse, the nails being 32 in number?

This question is evidently reduced to finding the sum of all the terms of the geometrical progression 1, 2, 4, 8, 16, etc. continued to the 32d term. Now, that last term is 2³¹; and, as we have already found 2²⁰=1048576, and 2¹⁰=1024, we shall have 2²⁰ · 2¹⁰ = 2³⁰ = 1073741824; and multiplying again by 2, the last term 2³¹=2147483648; doubling therefore this number, and subtracting unity from the product, the sum required becomes 4294967295 pence; which being reduced, we have 17895697l. 1s. 3d. for the price of the horse.1

512 Let the ratio now be 3, and let it be required to find the sum of the geometrical progression 1, 3, 9, 27, 81, 243, 729, consisting of 7 terms.

Calling the sum \(s\) as before, we have

\[s=1+3+9+27+81+243+729.\]

And multiplying by 3,

\[3=3+9+27+81+243+729+2187.\]

Then subtracting the former series from the latter, we have \(2s=2187-1=2186\): so that the double of the sum is 2186, and consequently the sum required is 1093.

513 In the same progression, let the number of terms be \(n\), and the sum \(s\); so that

\[s=1+3+3^2+3^3+3^4+\cdots+3^{n-1}.\]

If now we multiply by 3, we have

\[3s=3+3^2+3^3+3^4+\cdots+3^n.\]

Then subtracting from this expression the value of \(s\), as before, we shall have \(2s=3^n-1\); therefore \(s =\frac{3^n-1}{2}\). So that the sum required is found by multiplying the last term by 3, subtracting 1 from the product, and dividing the remainder by 2; as will appear, also, from the following particular cases:

1 \(\dfrac{(1\cdot 3)-1}{2}\) = 1
1 + 3 \(\dfrac{(3\cdot 3)-1}{2}\) = 4
1 + 3 + 9 \(\dfrac{(3\cdot 9)-1}{2}\) = 13
1 + 3 + 9 + 27 \(\dfrac{(3\cdot 27)-1}{2}\) = 40
1 + 3 + 9 + 27 + 81 \(\dfrac{(3\cdot 81)-1}{2}\) = 121.

514 Let us now suppose, generally, the first term to be \(a\), the ratio \(b\), the number of terms \(n\), and their sum \(s\), so that

\[s=a+ab+ab^2+ab^3+ab^4+\cdots+ab^{n-1}.\]

If we multiply by \(b\), we have

\[bs = ab+ab^2+ab^3+ab^4+ab^5+\cdots+ab^n,\]

and taking the difference between this and the above equation, there remains \((b-1)s=ab^n-a\); whence we easily deduce the sum required

\[s=\dfrac{ab^n-a)}{b-1}.\]

Consequently, the sum of any geometrical progression is found, by multiplying the last term by the ratio, or exponent of the progression, and dividing the difference between this product and the first term, by the difference between 1 and the ratio.

515 Let there be a geometrical progression of seven terms, of which the first is 3; and let the ratio be 2: we shall then have \(a = 3\), \(b = 2\), and \(n = 7\); therefore the last term is 3 · 2⁶, or 3 · 64, = 192; and the whole progression will be

3, 6, 12, 24, 48, 96, 192.

Farther, if we multiply the last term 192 by the ratio 2, we have 384; subtracting the first term, there remains 381; and dividing this by \(b - 1\), or by 1, we have 381 for the sum of the whole progression.

516 Again, let there be a geometrical progression of six terms, of which the first is 4; and let the ratio be ³⁄₂: then the progression is

4, 6, 9, ²⁷⁄₂, ⁸¹⁄₄, ²⁴³⁄₈.

If we multiply the last term by the ratio, we shall have ⁷²⁹⁄₁₆; and subtracting the first term = ⁶⁴⁄₁₆, the remainder is ⁶⁶⁵⁄₁₆; which, divided by \(b-1=\frac{1}{2}\), gives ⁶⁶⁵⁄₈ = 83⅛ for the sum of the series.

517 When the exponent is less than 1, and, conscquently, when the terms of the progression continually diminish, the sum of such a decreasing progression, carried on to infinity, may be accurately expressed.

For example, let the first term be 1, the ratio ½, and the sum \(s\), so that

\[s=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\textrm{etc.}\]

to infinity.

If we multiply by 2, we have

\[s=2+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\textrm{etc.}\]

to infinity: and, subtracting the preceding progression, there remains \(s = 2\) for the sum of the proposed infinite progression.

518 If the first term be 1, the ratio ⅓, and the sum \(s\); so that

\[s=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\textrm{etc.}\]

to infinity. Then multiplying the whole by 3, we have

\[3s=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\textrm{etc.}\]

and subtracting the value of \(s\), there remains \(2s=3\); wherefore the sum \(s=1\frac{1}{2}\).

519 Let there be a progression whose sum is \(s\), the first term 2, and the ratio ¾; so that

\[s=2+\frac{3}{2}+\frac{9}{8}+\frac{27}{32}+\frac{81}{128}+\textrm{etc.}\]

to infinity. Multiplying by ⁴⁄₃, we have

\[\frac{4}{3}s=\frac{8}{3}+2+\frac{3}{2}+\frac{9}{8}+\frac{27}{32}+\frac{81}{128}+\textrm{etc.};\]

and subtracting from this progression \(s\), there remains \(\frac{1}{3}s=\frac{8}{3}\): wherefore the sum required is 8.

520 If we suppose, in general, the first term to be \(a\), and the ratio of the progression to be \(\frac{b}{c}\), so that this fraction may be less than 1, and consequently \(c\) greater than \(b\); the sum of the progression, carried on to infinity, will be found thus:

Make

\[s=a+\frac{ab}{c}+\frac{ab^2}{c^2}+\frac{ab^3}{c^3}+\frac{ab^4}{c^4}+\textrm{etc.}\]

to infinity; and subtracting this equation from the preceding, there remains \(\left(1-\frac{b}{c}\right)s=a\). Consequently,

\[s=\dfrac{a}{1-\frac{b}{c}}=\frac{ac}{c-b},\]

by multiplynig both the numerator and denominator by \(c\).

The sum of the infinite geometrical progression proposed is, therefore, found by dividing the first term a by 1 minus the ratio, or by multiplying the first term \(a\) by the denominator of the ratio, and dividing the product by the same denominator diminished by the numerator of the ratio.

521 In the same manner we find the sums of progressions, the terms of which are alternately affected by the signs + and -. Suppose, for example,

\[s=a-\frac{ab}{c}+\frac{ab^2}{c}-\frac{ab^3}{c}+\frac{ab^4}{c}-\textrm{etc.}\]

And, adding this equation to the preceding, we obtain \(\left(1+\frac{b}{c}\right)s=a\): whence we deduce the sum required, \(s=\dfrac{a}{1+\frac{b}{c}}\), or \(s=\frac{ac}{c+b}\).

522 It is evident, therefore, that if the first term \(a=\frac{3}{5}\), and the ratio be ⅖, that is to say, \(b=2\), and \(c=5\), we shall find the sum of the progression

\(\frac{3}{5}+\frac{6}{25}+\frac{12}{125}+\frac{24}{625}+\textrm{etc.}=1\);

since, by subtracting the ratio from 1, there remains ⅗, and by dividing the first term by that remainder, the quotient is 1.

It is also evident, if the terms be alternately positive and negative, and the progression assume this form:

\[\frac{3}{5}-\frac{6}{25}+\frac{12}{125}-\frac{24}{625}+\textrm{etc.}\]

that the sum will be

\[\dfrac{a}{1+\frac{b}{c}} = \dfrac{\frac{3}{5}}{\frac{7}{5}} = \frac{3}{7}.\]

523 Again: let there be proposed the infinite progression,

\[\frac{3}{10}+\frac{3}{100}+\frac{3}{1000}+\frac{3}{10000}+\frac{3}{10000}+\frac{3}{100000}+\textrm{etc.}\]

The first term is here ³⁄₁₀, and the ratio is ⅒; therefore subtracting this last from 1, there remains ⁹⁄₁₀, and, if we divide the first term by this fraction, we have ⅓ for the sum of the given progression. So that taking only one term of the progression, namely ³⁄₁₀, the error would be ⅒.

And taking two terms, ³⁄₁₀ + ³⁄₁₀₀ = ³³⁄₁₀₀, there would still be wanting ¹⁄₁₀₀ to make the sura, which we have seen is ⁴⁄₃.

524 Let there now be given the infinite progression,

\[9+\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{9}{10000}+\textrm{etc.}\]

The first term is 9, and the ratio is ⅒. So that 1 minus the ratio is ⁹⁄₁₀; and \(\dfrac{9}{\frac{9}{10}}=10\), the sum required: which series is expressed by a decimal fraction, thus, 9.9999999 etc.

Editions

  1. Leonhard Euler. Elements of Algebra. Translated by Rev. John Hewlett. Third Edition. Longmans, Hurst, Rees, Orme, and Co. London. 1822.
  2. Leonhard Euler. Vollständige Anleitung zur Algebra. Mit den Zusätzen von Joseph Louis Lagrange. Herausgegeben von Heinrich Weber. B. G. Teubner. Leipzig and Berlin. 1911. Leonhardi Euleri Opera omnia. Series prima. Opera mathematica. Volumen primum.

  1. 1 pound (l.) = 20 shillings (s.), 1 shilling (s.) = 12 pence (d.), 1 pound (l.) = 240 pence (d.). ↩