In like manner (a + x)= ao+9a3x+36a2x2 + 84a3x3 +121a3x4 +121.a4x5+84a3x6 +36a2x2 +9ax2 + xo. It will also be observed that, in the case of (a + x)1o there is a middle term, 252a3x5; but in the case of (a+x) there is no middle term. In fact, it is plain that the expansion of (a+b)" contains n + 1. terms, for the expansion contains one term in which 6 does not appear, and also terms containing b, b2, b3 .. up to b. Hence, if n is even, n + 1 is odd, and the expansion has a middle term; if n is odd, n+1 is even, and the expansion is without a middle term. .... Now, when m and n are positive integers, we obtain from equation (I.) In accordance with the principle of the permanence of equivalent forms, we assume this to hold good, whatever m and n are, and then interpret the meaning we must assign to f(m) consistently with this assumption. And therefore, for all values of a positive or negative, integral or fractional, which is the Binomial Theorem, and we have proved it to hold good for all values of n positive or negative, fractional or integral. N.B.--The Binomial Theorem was discovered by Newton; it is very important, and is constantly used in almost every part of mathematics; it is therefore very necessary that the student be quite familiar with it. It will be well if he will carefully examine the following results, (1.) To show that We obtain this from equation (IV.) by writing a=1, and b—— X. write in the series (IV.) a=1 and b=1, and we obtain this result. Also, these fractions are all negative, .. if A, is positive, Ar+ 1 is negative, Ar+ 2 is positive, and so on. But if we neglect the consideration of the signs from what has been proved, it appears that A, 7 Ar + 1, Ar + 17 Ar + 2, &c. Hence, the part of the series beginning with A, x"; i. c. A, x2 + Ar + 1 + x2 + 1 + A‚ + 2 x2 + 2 + .... has its terms (if x be positive) alternately positive and negative, and is term by term less than Hence, provided x 1, the series, at all events, after a certain number of terms, converges. is true arithmetically. We may therefore apply this formula to extracting roots of numbers. Thus to extract the 5th root of 1-1. The fifth root of 1·1, is (1 + 1). a = (1+a-1)* = 1 + x (a − 1) + x(x − 1 (a− 1)2 + x(x − 1) (x — 2) .... 1.2 1)2+ 1.2.3 — 2) (a− 1)3 +........ (a) .... Now, it is plain that if we multiply the factors of the coefficients of (a — 1)2, (a—1)3, (a —1)1 . . together, we may re-arrange the series so that it shall become a2 = 1 + Ax + Bx2 + Cx3 + Dx2 + .......... where A, B, C, D, '. (b) contain (a1) and its powers in some determinate manner. For example, if we examine (a), we shall find that each term, after the first, contains the first power of a; viz., the second term contains x(a-1), the third contains hence the term involving the first power of x when (a) is re-arranged, must be (a−1)* +} &c. x * {( a − 1) — (a — 1)2 2 4 In like manner, we might find B. C. D. in series (6). Instead of doing so, we proceed as follows: since (b) is true for all values of x, we must have ay = 1+ Ay + By2 + Cy3 + Dy1· .. Now, a* Xay = a¤+y. .... .... 1+Ax + Bx2 + Сx3 + Dx1 +. .... (d) .... (e). And a+y=1+A (x+y)+B(x+y)a +C(x+y)3+....... =1+Ax+Bx2 + Сx3 + Dx1 + +Ay+2Bxy + 3Cx2y +4Dx3y+ +By2+3Cxy2+6Dx2y2+ (f) Now, from the early part of this article, it appears that there is one definite expansion of a*, and.. of ax + y. Hence (e) and (ƒ), which are each the expansion of a* + y, are not merely equal, but are actually identical; therefore they must be, term by term, Hence, a=1+ Ax+ +: + ····(V.), + · (V.), 1)3 _ (a − 1)* +.. (VI.). N.B. Suppose e to be such a number that— 4 1 = (e − 1) — (e — 1)2 (e− 1)3 __ (e — 1) 3 + 2 T |