and we shall have the series -1+1-1+1-1, &c., as before. And if we suppose a = 2, we shall have the series 262. In the same manner, by resolving the general fraction C a+b into an infinite series, we shall have, Let a 10, b = 1, and c = 11, and we have a+b 10+ 1 which If we consider only one term of this series, we have, which is too much by; if we take two terms, we have is too small by; if we take three terms, we have 1001 which is too much by Too, &c. 263. When there are more than two terms in the divisor, we may also continue the division to infinity in the same manner. Thus, if the fraction 1 were proposed, the infinite series, to which it is equal, would be found as follows: Eul. Alg. 11 1 1—a + aa = 1 + a — a3 — aa + aa + a2 — ao — a1o, &c. Here, if we make a 1, we have 1=1+1-1-1+1+1-1-1+1+1, &c. which series contains twice the series found above Now, as we have found this, it is not astonishing that we should find, or 1, for the value of that which we have just determined. Make a, and we shall then have the equation If we take the four leading terms of this series, we have 10, 264. The method, which we have explained, serves to resolve, generally, all fractions into infinite series; and, therefore, it is often found to be of the greatest utility. Further, it is remarkable, that an infinite series, though it never ceases, may have a determinate value. It may be added, that from this branch of mathematics inventions of the utmost importance have been derived, on which account the subject deserves to be studied with the greatest attention. CHAPTER VI. Of the Squares of Compound Quantities. 265. WHEN it is required to find the square of a compound quantity, we have only to multiply it by itself, and the product will be the square required. For example, the square of a +b is found in the following manner : a+b a+b aa+ab ab+b b aa+2ab+bb. 266. So that, when the root consists of two terms udded together, as a+b, the square comprehends, 1st, the square of each term, namely, a a and b b ; 2dly, twice the product of the two terms, namely, 2 a b. So that the sum a a +2ab+bb is the square of a +b. Let, for example, a = 10 and b = 3, that is to say, let it be required to find the square of 13, we shall have 100 +60 +9, or 169. 267. We may easily find, by means of this formula, the squares of numbers, however great, if we divide them into two parts. To find, for example, the square of 57, we consider that this number is 50+7; whence we conclude that its square is =2500 +700 +49 3249. 268. Hence it is evident, that the square of a +1 will be aa+2a+1: now since the square of a is a a, we find the square a+1 by adding to that 2a+1; and it must be observed, that this 2a+1 is the sum of the two roots a and a + 1. Thus, as the square of 10 is 100, that of 11 will be 100 +21. The square of 57 being 3249, that of 58 is 3249 +115= 3364. The square of 59 = 3364 +117 = 3481; the square of 603481119 = 3600, &c. 269. The square of a compound quantity, as a+b, is represented in this manner: (a+b)2. We have then (a+b)2 = aa+2ab+bb, whence we deduce the following equations: (a + 1)2 = a a + 2 a + 1 ; (a + 2)2 = a a + 4 a + 4; 6a+ (a + 3)2=aa + 6 a + 9; (a + 4)2=aa+8a+ 16; &c. 270. If the root is ab, the square of it is a a—2ab+bb, which contains also the squares of the two terms, but in such a manner that we must take from their sum twice the product of those two terms. 271. Since we have the equation (a—b)2= aa—2ab+bb, we shall have (a—1)3 =a a—2a+1. The square of a-1 is found, therefore, by subtracting from a a the sum of the two roots a and a-1, namely, 2 a-1. Let, for example, a = 50, we have a a=2500, and a-149: then 492 = 2500 - 992401. 272. What we have said may be also confirmed and illustrated by fractions. For if we take as the root+ (which make 1) the squares will be: 9 4 23 + 1/3 + 1/3 = 3, that is 1, Further, the square of — } (or of 3) will be 1 = 273. When the root consists of a greater number of terms, the method of determining the square is the same. for example, the square of a+b+c. Let us find, |