To insert m geometric, means between a and b. Here we are required to form a geometric series, of which the first and last terms, a and b, are given, and the number of terms = m + 2; in order, then, to determine the series, we must find the common ratio. Eliminating S by equations (1) and (2), a+am+1 bm+1 +am+16m+1 +...+am+16m+1 +am+16m+1+o 182. To find the sum of an infinite series, decreasing in geometrical progression. We have already found, that the sum of n terms of a decreasing geometrical series, is, S= which may be put under the form, n Since is a fraction, ę" is less than unity, and the greater the number n, the smaller will be the quantity "; if, therefore, we take a very great number of αρ" terms of a decreasing series, the quantity ę", and, consequently, the term and if we take n greater than any will be very small in comparison with a assignable number, or make n = ∞, then " will be smaller than any assignable number, and therefore may be considered = 0, and the second term in the above expression will vanish. Hence, we may conclude, that the sum of an infinite series, decreasing in geometrical progression, is, terms approaches, and the above expression will approach more or less nearly to perfect accuracy, according as the number of terms be greater or smaller. Thus, let it be required to find the sum of the infinite series Thus, if n = terms of the above series, is determined by the quantity, n = 6, then, 3 (-)° as the sum of 5 terms of the above series, the amount 1 would be too great by 162 as the sum of 6 terms, the amount would be too great by 486' and so on. HARMONICAL PROGRESSION. 183. A series of quantities are said to be in harmonical progression, when, if any three consecutive terms be taken, the first is to the third as the difference of the first and second to the difference of the second and third. 184. The reciprocals of a series of terms in harmonical progression are in arithmetical progression. Let a, b, c, d, e, f, . . . ... be a series in harmonical progression, Then, by definition, ac bc bc bd bd dc cd a : c : : a − b : b—c;b:d :: b—c: c—d; c:e::c—d: d—e; &c. ab. ab ac — ac ac—bc, bc-bdbd-dc, cd -ce ce ed, &c. from which it appears, that the quantities arithmetical progression. To insert m harmonic means between a and b. 증, &c. are in Since the reciprocals of quantities in harmonical progression are in arithmetical progression, let us insert m arithmetic means between 185. The Permutations of any number of quantities signify the changes which these quantities may undergo with respect to their order. b c an Thus, if we take the quantities a, b, c; then, a b c, a c b, b a c, ca b, c ba, are the permutations of these three quantities taken all together; ab, ac, ba, bc, ca, cb, are the permutations of these quantities taken two and two; a, b, c, are the permutation of these quantities taken singly, or one and one, &c. The problem which we propose to resolve is, 186. To find the number of the permutations of n quantities, taken P together. Let a, b, c, d, ......... k, be the n quantities. and P The number of the permutation of these n quantities taken singly, or one and one, is manifestly n. The number of the permutations of these n quantities, taken two and two together, will be n (n-1). For since there are n quantities, That is, (n - 1) permutations of the n quantities taken two and two, in which a stands first. Reasoning in the same manner for b, we shall have (n —1) permutations of the n quantities taken two and two, in which b stands first, and so on for each of the n quantities in succession, hence the whole number of permutations will be The number of the permutations of ʼn quantities taken three and three together is n (n − 1) (n-2). For since there are n quantities, if we remove a there will remain (n 1) quantities; but, by the last case, writing (n 1) for n, the number of the permutations of (n − 1) quantities taken two and two is (n (n — 2); writing a before each of these (n − 1) (n - 2) permutations, we shall have (n − 1) (n — 2) permutations of the n quantities taken three and three, in which a stands first. Reasoning in the same manner for b, we shall have 1) (n 1) (n 2) permutations of the n quantities taken three and three in which b stands first, and so on for each of the n quantities in succession, hence the whole number of permutations will be In like manner we can prove that the number of permutations of n quantities taken four and four will be Upon examining the above results, we readily perceive-that a certain relation exists between the numerical part of the expressions, and the class of permutations to which they correspond. Thus the number of permutations of n quantities, taken two and two, is n (n − 1) which may be written under the form n (n -2+1) Taken three and three, it is n (n − 1) (n — 2) which may be written under the form n (n − 1) (n − 3 + 1) 3) which may be written under the form n (n − 1) Hence from analogy we may conclude, that the number of permutations of n things, taken and p together, will be In order to demonstrate this, we shall employ the same species of proof already exemplified in (Arts. 39 and 89), and show that, if the above law be assumed to hold good for any one class of permutations, it must necessarily hold good for the class next superior. Let us suppose, then, that the expression for the number of the permutations of n quantities taken (p - 1) and (p · 1) together is It is required to prove that the expression for the number of the permutations of n quantities, taken p and p together, will be Remove a one of the n quantities a, b, c, d................................k, then by the expression (A), writing (n 1) for n, the number of the permutations of the (n − 1) quantities b, c, d........................k, taken (p-1) and (p— 1, will be - (n-1) (n-2) (n-3).......... {(n - 1) — (p − 1) + 1} p+1) permu tations of the n quantities, in which a stands first. Reasoning in the same manner |