6. What must be the annual increase of population in any country, that it may double itself every century? Here m=2 } 1: 1 logm log 2 301030 003010 and th. Ρ the arithmetic means between 128. Supposing that a census of the whole population of a country is taken every n years, and that it is found to have increased per cent. during that interval, then if P represents the amount of the population at the commencement of the n years, P+ will represent the amount of the population at the end of the 100 1 If the annual increase be then (by Art. 127) the p n years. n n ΠΡ 100 = P(1+1) '; ; hence P(1+ P(1+ `.. n · log (1 +1)—log (100+7)—log 100—log (100 + π) —2, since log 100—2, and log(1+1)—— {log (100+~)—2}. Substitute this value of log(1+in the expression the number of years in which the population of a country will be increased m times, if it goes on increasing at the same rate as it has done for the last n years preceding the period at which the census is taken. 129. If the census be taken every ten years, and the period of doubling be required, then n=10, m=2, and the foregoing expression becomes By substituting in it for the particular value of the per centage, the following table exhibits the corresponding period of doubling. See index. ON Figurate and Polygonal Numbers. 130. Figurate numbers are such as arise from taking the successive sums of the series of the natural numbers 1, 2, 3, 4, &c., and then the successive sums of these last, and so on; and Polygonal numbers are those which are formed of the successive sums of the terms of any arithmetical progression, beginning with unity; each of them being usually divided into orders, according to the scale of their generation, which, as far as regards those of the first class, may be shown as follows: Order. Figurate Numbers. 1, 2, 3, 4, 5, 6, &c. General Terms. Nat. ser ies. n 1st order. 21, &c. n(n+1) 1.2 2d order. 1, 4, 10, 20, 35, 56, &c. 3d order. 1, 5, 15, 35, 70, 126, &c. n(n+1)(n+2) n(n+1)(n+2)(n+3) where it is to be observed, that the general terms here given are so called, because if 1, 2, 3, &c. be respectively substituted, in each of them, for n, we shall obtain the several terms of the series. And if, instead of the natural numbers 1, 2, 3, 4, &c. which give, by their addition, triangular numbers, an arithmetical series be taken, the common difference of which is 2, the sum of its successive terms will be the series of square numbers. If the common difference be 3, the series will be pentagonal numbers; if 4, hexagonal; and so on. Thus. Arith. Series. Or. Polygonal Numbers. Gen. Terms. 1, 2, 3, 4, &c. | 1 | Tri. 1, 3, 6, 10, 15, 21, &c. | {{n(n+1)}, 1, 3, 5, 7, &c. | 2 | Sqrs. 1, 4, 9, 16, 25, 36, &c. {n(2n+1)} 1, 4, 7, 10, &c. 3 Pent. 1, 5, 12, 22, 35, 51, &c. {n(3n-1)} 1, 5, 9, 13, &c. | 4 | Hex. 1, 6, 15, 28, 45, 66, &c. | {{n(4n—2)} where the number denoting any order is the common difference of the arithmetical series from which the polygonal numbers, belonging to that order, are generated. In like manner, if we take the successive sums of the several polygonal numbers thus formed, and then the successive sums of these last, and so on, a great variety of other orders of this kind may be obtained. Hence, also, in general, if n be made to denote the number of terms of the series, a figurate number of any order m, which will consist of m + 1 factors, may be expressed by the following formula: And supposing n to be the number of terms of the series, as be fore, a polygonal number of the order m-2, or one of which the number of sides of the polygon is denoted by m, may be expressed (m—2) n2—(m—4)n, by so that figurate Nos. of any order, may be also determined without computing those of the preceding or ders, by taking as many factors, in the first of these formulæ, as is denoted by the number of the given order plus 1, and making ʼn equal to the term that is to be found. 2 And a polygonal number of any order, or number of sides, may be ascertained from the second of these formulæ by substituting the number denoting that order for m―2, or the number of sides of the polygon, for m, and taking n equal to the term required. 1. Required the 15th term of the 1st order of figurate numbers 1, 3, 6, 10, 15, &c. Here the number of factors being two, and n=15, we shall have, by the first formula, n ̧n+115(15+1) 15×16 2 2 2-15X8-120, the term required. 2. Find the 20th term of the 4th order of figurate numbers. Here, the number of factors being 5, and n=20, we shall have ‚n+1n+2n+3n+4_20 21 22 23 24 X- X X X n 3 52504, the ans. term required. 5 4 5 3. It is required to find the 12th term of the fifth order of poly. gonal numbers, being those called heptagonal, or such as would be represented by a figure of seven sides. Here m—2 being =5, or m=7, and n=12, we shall have, by the 2d formula {(m—2)n2—(m—2)n}={{(7—2)×144—(7—4)} X12=5X72-3x6-360-18-342, the term required. 1. Find the 13th term of the 8th order of figurate numbers. 2. It is required to find the 36th term of that order of polygonal numbers which is denoted by a figure of twenty-five sides. 3. It is required to find the first seven terms of the 6th order of figurate numbers. Ans. 293930. Ans. 14526 Ans. 3011 4. It is required to find the first twelve terms of the order of polygonal numbers called nonagonal, or such as are denoted by a figure of nine sides. Ans. On Interest and Annuities. 131. Interest is the consideration paid for the use or forbear ance of the payment of money. Rate of interest is the consideration paid for the use of a certain sum for a certain time. Thus, $5 per cent. per annum means, that $5 are to be paid by the borrower to the lender for the use of $100 for a year. When the interest of the principal, or sum lent, is taken, it is called simple interest; but if the interest, as soon as it becomes due, be considered as principal, and interest be charged upon the whole, it is called compound interest as before. 132. Amount is the whole sum due at the expiration of any time, principal and interest together. Discount is the abatement made for the payment of money before it becomes due. 133. The present worth of any debt due some time hence is such a sum as, being put out to interest for that time, will amount to the debt.* interest. = 134. To find the amount of a given sum in any time, at simple · Let p principal or money lent, r = interest of $1 for a year, n = time for which interest is required, m = amount. Now it is evident that the interest of a given sum, at a given Yr. Yrs. rate, must be proportional to the time. Hence 1:n : : r : nr = interest of $1 for n years, and the interest of $P being P times as great, ... Prn the whole interest of $P for (n) years at the proposed rate. Now amount = principal therefore MP+ Prn = interest; In this simple equation, any three of the quantities P, n, r, M, being given, the fourth may be found. 5 100 -- per ct. 1. Find the amount of $280.5 for 3.years, 148 days, at 5 per annum. Here P 280.5, r= = —$·05, n=3.40547 years. .. amount = 280.5 1.1702735) 328-26171675 $328 2. Find the principal which, being put out to interest for 189 days at 4 per cent. will amount to $200. Ans.$196. nearly. 135. To find the amount of an anniity or pension, left unpaid any number of years, allowing simple interest upon each sum from the time it becomes due. Let A be the annuity, then, at the expiration of one year, A becomes due, and at the end of the second year the interest of the first annuity is Ar; also at the end of this year the principal sum due is 2A, and its interest at the end of the third year is 2Ar, &c. hence, at the end of n years, the sum due is .·. nA+rA+2rA+&c.....(n-1)rA, .. M=nA+}{n(m—1)rA.} When the pensions are payable half yearly, MnA +m(2n-1)Ar, and if pensions payable quarterly Hence true discount is charged on any sum when the difference between the sum and its present worth is taken. MnA+2n (4n-1),Ar. 1. If an annuity of $70 be forborne five years, what will it amount to at 5 per cent.? (n-1) 5X4 [$385. rA=5x70 + X·05×70—350 +35= = 2. Required the same as in the last question, supposing halfyearly payments to be made. MnA+n(2n-1) Ar-350+ 5X9X70X.0125 $389.375. If the payments are made quarterly, the amount will be $391.561 136. To find the present value of an annuity to continue a certain number of years, allowing simple interest. Let the present value, then if x and the annuity at the same rate of interest amount to the same sum, they are of equal value. Now, x+nrx= amount of x in n years at rate r; also rA amount of the annuity. ..x+nrx=nA+ nA+ n• (n—1) 2 1. What is a pension of $30 per annum, for 5 years, worth in ready money, at 4 per cent.? x= 300+30X30X 045 2+45 $133·469=$133.47 2. What is the present worth of $50 per annum, payable quar terly for 6 years, at 5 per cent? 2nA+n (n-1)rA Ans. $263.94 if n be infinite, x is In the equation x = infinite, which leads us to suppose that for a limited annuity to continue for ever, an infinitely great sum ought to be paid; a conclusion which shews the necessity of estimating the value of an annuity upon different principles. On Compound Interest. 137. Interest is a certain sum paid for the use of money for any stated period, and when the interest of this money is regularly received, the money, or principal, is said to be at simple interest. But when, instead of being regularly received, it is allowed to go to the increase of the principal, then the interest of the whole is called compound interest. 138. An annuity is a yearly income, or pension. the 139. The present value of an annuity is that sum which, if put out at compound interest, shall amount to sufficient to pay annuity at the time it becomes due. 140. Let P be the principal, or sum put out to compound interest, r the fraction which expresses the rate of interest per cent.* |