The 307. The method of raising a binomial to any power by repeated multiplication has been before laid down in Art. 140. same thing may be done much more expeditiously by the following general rule, which is called the Binomial Theorem. Let aa be the binomial; its nth power is Where the index of a, beginning from n, is diminished by unity, and the index of a, beginning from 0, is increased by unity, in every succeeding term. Also the coefficient of each term is found by multiplying the coefficient of the preceding term by the index of a in that term, and dividing by the index of a increased by unity. =x+6ax+15a2x2+20a3x3+15a x2+6ax+a®. 308. To investigate the Binomial Theorem for a positive integral By actual multiplication it appears that index. (x+a)(x+b) = x2+(a+b)x+ab, (x+a)(x+b)(x+c)=x3+(a+b+c)x2+(ab+ac+bc)x+abc, (x+a)(x+b)(x+c)(x+d)=x*+(a+b+c+d)x3 +(ab+ac+bc+ad+bd+cd)x2 and the same law of formation of the continued product is observed to hold whatever be the number of binomial factors, x+a, x+b, x+c, &c., actually multiplied together, viz. that it is composed of a descending series of powers of x, the index of the highest being the number of factors, and the other indices decreasing by 1 in each succeeding term. Also the coefficient of the first term is 1; of the second the sum of the quantities a, b, c, &c.; of the third the sum of the products of every two; of the fourth the sum of the products of every three; and so on; of the last the product of all the n quantities a, b, c, &c. Suppose, then, this law to hold for n binomial factors, x+a, x+b, x+C,................, x+k; so that (x+a)(x+l)(x+c)... (x + k) = x2+AxTM-1+Bx2-2+Cx"¬3+...+K, B=ab+ac+be+... C=abc+acd+... &c.=&c. K=abcd...k; introducing a new factor, x+l, we have (x+a)(x+b)(x+c)...(x+k)(x+l) = x2+'+ ( A +l)x" + (B+lA)x^~'+...+Kl. so that, if the law above described holds when n binomial factors are multiplied together, the same law is proved to hold for n+1 factors. But it has been shewn to hold up to 4 factors, therefore it is true for 5; and, if for 5, then also for 6; and so on, generally, for any number whatever. Now, let a=b=c=&c., Also (x+a)(x+b)(x+c)...(x+k) becomes (x+a)"; 309. Having given the Binomial Theorem for a positive integral index, to prove it, when the index is either fractional or negative. [EULER'S PROOF.] m(m-1) +&c. be represented for all values Let the series 1+mx+ of m, whether positive, or negative, integral, or fractional, by the symbol f(m); then it has been shewn that, when m is a positive integer, f(m)=(1+x)TM. It remains to prove that this equation is also true when m is either fractional or negative. therefore, by multiplication, f(m).f(n)=the product of the two series, which will evidently be a series of the form 1+ax+bx2+cx2+ &c., ascending regularly by the integral powers of x, the coefficients a, b, c, &c. being different combinations of m and n. Now, although by changing the values of m and n the values of a, b, c, &c. are altered, yet their forms, that is, the manner in which m and n enter the series will remain the same. Whatever, therefore, be the forms of a, b, c, &c. when m and n are positive integers, the same will they be when m and n are fractional or negative. But in the former case These then are the forms in which m and n appear in the product when they are positive integers; and therefore they appear in these same forms, whatever be their values: i.e. whether m and n be positive or negative, integral or fractional, the multiplication of f(m) by f(n) always produces the series 1+(m+n)x+ (m+n)(m+n−1) x2+ &c. 1 2 which by the notation is represented by f(m+n); .. universally, f(m).ƒ(n)=f(m+n). Since, then, this last equation is true whatever be the values of m and n, for n write n+p, and we have f(m+n+p)=f(m).ƒ(n+p)=f(m).f(n).f(p); and proceeding similarly, we have generally f(m+n+p+&c.)=ƒ(m).ƒ(n).ƒ(p). &c. to any number of terms. but ƒ(h)=(1+x)", because h is a positive integer: which proves the Theorem for a fractional positive index. Again, ··ƒ(m).ƒ(n)=ƒ(m+n) for all values of m and n, let n=-m, then f(m).ƒ(−m)=f(m—m)=ƒ(0), = 1, . the assumed series becomes 1 when m=0; −m(−m−1) x2+... or (1+x)—"= f(−m)=1+(-m)x+1.2 which proves the Theorem for a negative index, integral or fractional. 310. To prove the Binomial Theorem for any index whatever. [GRIF FITH'S PROOF.*] Let a, denote a(a-1). 1.2 ; β. ; * This proof, unlike most others, is dependent only upon the most simple elementary rules of Algebra. ED. Now the product of the two series 1+ax+açx2+a ̧x3 +...+at+............(1) 1+ßx+ß ̧x2+ß ̧æ3+...+ß,x +(2) by actual multiplication, is 1+(a+6)x+terms of x2, x3, ...x”, ... the coefft. of being a,+a,-ß+a‚—‚ß2+¤‚¬3ß ̧+...+aßrı+ ßr, and ................. x' -1... α-+α-2ß+Ⴌ3ß2+a‚ß2+...+aß12+ẞr-1 raẞ (r−1)α,ß + a‚‚ß =(a−r+2)a,_2ß +ßa,-1, .. adding, and observing that there is a pair of similar terms in every two lines, we have rxcoeff. of x=(a+ß−r−1)(a,+a,ß+a‚‚ß2+...+ßm-1); and writing 2, 3, 4, ... successively for r, the product of the series (1) and (2) becomes therefore calling a the characteristic' of series (1), and ẞ of (2), it is proved that the product of two such series is a similar series with the sum of their characteristics' for its characteristic.' Hence also this product multiplied by another such series, that is, the product of three such series, will be an exactly similar series with the sum of their characteristics' for its 'characteristic': and so on for any number whatever of such series. 1. Let the number of series be n, and the 'characteristic' of each 1; then all the series being alike, and the sum of their characteristics' n, we have |