189. Let the sum of all the terms of the series a bed....: k : l, be denoted by S; that is, let S=a+b+c+d+....+k+l. Then, if all the terms are multiplied by q, qS=qa+qb+qc+qd+......+qk+gl. But qa=b, qb=c, qc=d, .....qk=l; therefore, by substitution, qS=b+c+d+... ......+1+ql. From this equality subtract S=a+b+c+d+ ...... ...qS-S=ql-a, or (q−1)S=ql—a, and S-l-a - 3. +1; Whence the sum of the terms of a geometrical progression is obtained by multiplying the last term by the ratio, subtracting the first term from the product, and dividing the remainder by the ratio, less 1. 190. If is replaced by its value, aq"-1, formula 3 becomes S= _a(q"—1) 4. In formula 4, suppose 1st q>1, which is the case of increasing progressions. In this case the quantity q" is great in proportion as n is great. There is no limit even which it may not exceed if a sufficiently considerable value is given n(n-1) to n. For making q=1+a, q′′=(1+a)"=1+na+ ́a2+, &c. .'. 1.2 q">1+na. Now, however small the quantity a, it is evident that, by taking n sufficiently great, 1+na can be made to exceed any assigned quantity however great; therefore, à fortiori, this must be true of q”. Hence when q>1, by taking a sufficient number of terms of the progression a, aq, aq2 .. S can be made to exceed any assigned quantity. .... Suppose, 2d, q<1, which is the case of decreasing progressions. If the signs of the numerator and denominator are changed, formula 4 may be written thus ; a aq" -q l-q In this expression the quantity q" is small in proportion as n is great; it can even be rendered less than any assigned quantity. To prove this, let 1 ß being greater than 1; 1 then q′′= ; now if n is made greater B" and greater, ß" may be augmented to infinity, and by consequence reduced to zero; by which means is also reduced to zero. aq" 1-q or q", Hence it is evident that the sum S must increase as n is increased, and that it can never a exceed the limit S=1-q from which it may, however, be made to differ in defect by less than any assigned quantity. That is to say, the sum of the terms of a decreasing geometrical progression, continued to infinity, is equal to the quotient of the first term divided by unity, less the constant ratio. To obtain the sum of the terms of the series 1+1+}+&+, &c. make a=1 and q=4; then, 1 -- Suppose, 3d, q=1. In this case formula 4 gives S=g, an indeterminate result. But the indetermination is apparent only, for if q"-1 is divided by q-1, the expression is reduced to S=a (q′′-1+q"−2+ +q+1), which for the hypothesis q=1 gives S=an. That this must be the result is evident à priori, for the ratio being 1 each of the n terms of the progression is equal to the first term a. The division of q-1 by q-1 reproduces, in a progression of which formula 4 represents the sum. _a(1-q) Writing the formula thus, S=" contrary order, that and effecting the division of 1—q′′ by 1-q, the terms of the progression are obtained in the original order. In like manner formula 5, which expresses the sum of the terms of a decreasing progression prolonged to infinity, may be made to reproduce the progression a: aq: aq2, &c.: Dividing a by 1-q, as follows, it appears that according as the division is ended at the 1st, the 2d, the 3d.... remainder, it is necessary, in order to complete the corresponding aq aq2 aq3 quotients, to add to them respectively the fractions 1-q' 1-q, 1-q' Now q being less than 1, these fractions have decreasing values; and since the division may be continued until the exponent of q in the remainder is greater than any assigned number, and the value of the numerator of the complementary fraction by consequence less than any assigned number, it follows that, supposing the terms of the quotient continued indefinitely, the complementary fraction or remainder ought to be regarded as zero, therefore the series of these terms continued to infinity is the exact value of the quotient. This conclusion is legitimate only in the hypothesis q<1. If q>1, the complementary fractions or remainders, instead of approximating to zero, increase to - 8. 191. In geometrical as in arithmetical progression all the questions which can be proposed are reducible to ten. The ten questions, with their solu tions (which are deduced from the two equations l=aq", s=2—4), 1st. S= , are, Given a, q, n, to find l, S?......Ans. l=aq-1; Sql_a_a(qn—1) == q- 9-1 l(g"-1) ; S= (1). "VF-"Va" a 4th. Given n, q, S, to find a, l? ....................Ans. a= 5th. Given a, n, S, to find l, q? S(q-1) S Ans. q+q+ 6th. Given 1, n, S, to find a, q? ... Ans. ql-a 7th. Given a, l, q, to find S, n?...Ans. S= ; n=1+ -1 S-a 8th. Given a, l, S, to find q, n ?........Ans. q=5—¡ ; n=1+; 9th. Given a, q, S, to find l, n? 10th. Given l, q, S, to find a, n? log. l-log. a log. q log. l-log.a log. q log. l-log. a log. q Ans. 1 a+S(q-1) ; n=1+ log. l-log. a log. 9 Ans. a=lq-S(q−1); n=1+' Remarks. In case 5, 7 is eliminated between the equations l=aqi-2 and ql-a 9-1' S= the result is an equation in q of the degree n-1, which, unless in the particular case n=3, cannot be resolved by any of the preceding rules. The value of the ratio q being determined, is found from the equation laq”— In case 6, instead of q is made the unknown quantity. The preceding q remark respecting case 5 applies to case 6 also. gl-a In cases 7, 8, 9, 10, the equation S= 9-1 gives each of the quantities a, 1, q, S, in terms of the three others. The only difficulty, therefore, in these cases lies in the determination of the unknown quantity n from the equation =aq, in which n is an exponent. The solution given depends on certain properties of logarithms, for which see Section X. Examples: 1st. 2d. Find the 7th term of the series 1, 4, 16, &c. ?..........................Ans. 4096. Find the 8th term of the series 81, -27, 9, &c. ?........Ans. —73d. Find the 5th term of the series 4, -12, 36, &c. ?......... Ans. 324. 4th. Find the 6th term of the series 3, 1, 7, &c. ?......... Ans. 5th. Find the sum of 3+6+12+, &c. to 6 terms?. .........Ans. 189. 6th. Find the sum of 1-4+16-, &c. to 7 terms?............Ans. 3277. 7th. Find the sum of +++, &c. to 8 terms ?.............Ans. 85 8th. Find the sum of 3-4+, &c. to 5 terms ?................................Ans. 2247. 9th. Find the sum of the series 1+1++, &c., continued to infinity? Ans. 2. 10th. Find the sum of the infinite series 4+2+1+, &c. ?.......... Ans. 8. 11th. Find the sum of the infinite series 1-+-, &c. ?.......................Ans. §. 12th. Find the sum of the infinite series -3+18-+, &c.? Ans. -2. 13th. Insert 3 geometrical means between 2 and 32 ?.Ans. +4, 8, ±16. 14th. Insert 4 geometrical means between - and 34 ? Ans. —,—13. 15th. Insert 3 geometrical means between and 128 ?Ans. +2, 8, +32. 16th. What is the sum of the infinite geometrical progression b2 63 64 + 2 ............. Ans. a+b. a3 a a-b+ a &c. ?.......... OF THE NUMBER OF BALLS IN A PILE. 192. If all the terms of a geometrical progression a aq aq aq3 : . . . . aq", are raised to the same power, m, the result is the series, a2 which is a geometrical progression, of which the first term is a", the ratio q", and the number of terms n. Whence, if all the terms of a geometrical progression are raised to the same power, the result is another geometrical progression. a. Next, if a, a+d, a+28, a+36, are consecutive terms of an arithmetical progression whose ratio is d, and if each of these terms is raised to the power m, and the differences of the mth powers of the consecutive terms are taken; since the differences are not constant; therefore the same powers of the terms of an arithmetical progression are not in arithmetical progression. If, however, all the terms of an arithmetical progression are raised to the same power, m (m being integer and positive), the sum of the new series can be found. Leta. b. c. d.. kl, be any arithmetical progression, and let the common ratio be denoted by d. Then ba+; c=b+d; Raising these equalities to the power m+1, .1=k+d. (m+1)m am-152+, &c. bTM+1=a+1+(m+1)aTMd+ 1.2 cm+1=bm+1+(m+1)bTM8+ (m+1)m bm-162+, &c. Adding these equalities member by member, suppressing the common terms bTM+1, c”+1, &c., and transposing aTM+1, 7m+1 — am + 1 = (m+1)ε(aTM+bTM+ .... +k") To abridge this expression, let a+b+c+ +k+1=S,; m+1 .... a2+b2+c2+ +k2+l2=S2 ; .... The value of S, deduced from this series is 7m+1—am+1 m m(m-1) 23 S=1"+ The law of the unwritten terms is sufficiently apparent, and the series must evidently end with the term preceding that which contains the factor m-m, or 0. By formula 1 the sum S can be found when the inferior sums are known; for, making m=0, the formula gives S.; making m=1, it gives S1; making m=2, it gives S, and so on to the sum which is required. ...... b. If the progression +a. a+8. a+28. a+38 . . . . . . . is replaced by +1.2.3.4...... N (or the series of numbers from 1 to Ñ), a=1, =1, 1=N, and formula 1 becomes N+1-1 m S=N"+ m+1 If m=0, formula 2 becomes m(m-1) (Sm-2-N-)-, &c. - 2 2.3 S=N+ ==1+ 0+1 Formula 3 expresses the sum of 1°+2°+3o+ 1 2 3 1+ 2+ 3+ ..... N + N; or 1+1+1+1+... Example: If m=0 and N=10, S=N=10. Formula 4 expresses the sum of 1+2+3+4+ - 6 10(10+1)_110 2 -=55. Example: If m=1 and N=10, S1= 2 Formula 5 expresses the sum of 12+22+3+42+ . . . . . +N2, or the sum of the squares of the series of natural numbers from 1 to N inclusive. If m=2 and N=10, S= Formula 6 expresses the sum of 13+23+33 +43 + ..... +N3, or the sum of the cubes of the series of natural numbers from 1 to N inclusive. 10 (10+1) 100×121 Example: If m=3 and N=10, S= 4 = 4 -=3025. 193. In arsenals balls of the same size are arranged in piles; and it is by means of these formulæ (Art. 192) that the number of balls in a pile is computed. These piles are either, 1st, triangular; or, 2d, quadrangular; or, 3d, rectangular. |