or a a(a−1)................(a−q+1) b3y3, (where q+ß=a, or p+q+ß=m), 1. 2 ...... 4 ; so that the general term of the multinomial becomes Again, if d+&c.=x, y3 = (x+c), of which the general term, expressed by the r+1th, is or B. Try r c'xy, where (r+y=ß, or p+q+r+y=m), (y-1)...3.2.1 r 1 2 -c'a, (y being a pos. integer), ; so that the general term of the multinomial becomes and so on, until the terms of the multinomial are exhausted. Hence the general term required is p being fractional or negative when m is fractional or negative, but q, r, s, &c. always positive integers. COR. If m be a positive integer, then, since p is a positive integer, the expression for the general term may be written The last result may also be arrived at by the following method, assuming the index a positive integer: 328. To expand (a+b+c+d+&c.)TM, when m is a positive integer. and if €2.7182818, expanding by the Exponential Theorem, (Art. 325). 1+(a+b+c+d+&c.) + (a+b+c+d+&c.)° ~ +..... Now, as this operation merely exhibits the same quantity expanded in two different ways by the same theorem, the corresponding terms, that is, the terms involving the same powers of a will be equal to each other; therefore equating the coefficient of a" on the one side with the coefficient of " on the other, and observing that each separate term on this side of the equation which involves a" will be the product of as many terms as there are series to be multiplied, one of which is taken out of each series, and will therefore be of the form comes abc. &c. P. q. r. &c. abc.&c. P. q.r.&c. m P.q.r.&c. a2bc. &c.t COR. 1. If q+r+s+&c.=T, then p=m-, and the general term be- a-bic. &c. which form is sometimes found more convenient. Ο COR. 2. If it be required to expand (a+a,x+a ̧x2+azx3+&c.)TM, the general term may be obtained from that of (a+b+c+d+&c.)" by writing a ̧, a ̧x, a‚x3, ɑ ̧x, &c. in place of a, b, c, d, &c. respectively, by which it becomes and all the terms of the expansion may be found as before, by giving p, q, r, s, &c. all possible values which the condition p+q+r+s+&c.=m admits of. Also any particular term involving a proposed power of a, as a", will be found by taking the sum of the values of this general term, when P, q, r, s, &c. are made to assume all the values, which satisfy the two equations p+q+r+s+&c.=m, and q+2r+3s+&c. = n. ⚫ Σ stands for the expression "the sum of all the quantities of the form of." + The proof here given of the Multinomial Theorem extends only to the case of positive integral indices, for by the Exponential Theorem m cannot be any thing but a positive integer. But if the Multinomial be deduced from the Binomial Theorem, (as in Art. 327) then since the latter is proved for fractional and negative indices, the former is also proved to hold for such indices. COR. 3. Assuming the Theorem for a positive integral index, it may be proved for a fractional or negative index thus: Let b+c+d+&c.=x, then (a+b+c+d+&c.)TM=(a+x)", of which the general term, expressed by the a+1th, is m(m-1)... (p+1) uxa, where p+a=m; and since a is a positive integer, by what has been proved the general term of xa, or (b+c+d+&c.)a, is [a. b c &c. q r .. term required=m(m-1)...(p+1).a". ..&c. Ex. 1. Required the term in the expansion of (a-b-c) which involves a b3c3. Here m=7, m-n=2, q=3, r=2; 7.6.5.4.3 .. the term required: = . a3. (-b)3. (−c)3, Ex. 2. 1.2.3.1.2 =- -210a2b3c2. Required the term in the expansion of (a + bx+cx2+dx3)• which involves Ã3. remains to find all the To do this, it is most highest or the lowest, where p+q+r+s=4, and q+2r+3s=8; and it values of p, q, r, s which satisfy these equations. convenient to take in order, beginning with the the several values of q, r, and s, which satisfy the latter, and reject those which are inconsistent with the former, equation. And it is most advantageous to begin by assigning values to that quantity which has the greatest coefficient, i.e. in this case s, and to take them in descending order of magnitude. Thus we see from the second equation that s cannot be greater than 2: i.e. it may have the values 2, 1, 0: let s=2; then q+2r=2: whence r=1 or 0, and q=0 or 2. Next, let s=1; then q+2r=5; and r=2, 1, or 0; q=1, 3, or 5. Lastly, let s=0; then q+2r=8: and r=4, 3, 2, 1, or 0; q=0, 2, 4, 6, 8. Of these values, only so many must be taken as will satisfy the first equation, when we shall have to reject all except the underwritten: where the simultaneous values of the quantities are written from left to right. Ex. 3. Find the coefficient of x in the expansion of (a-bx+cx)'. .. coeff. required=[12. + [4.8 5.6 6.4.2 7.2.3 8.4) =495a*b*+5544a b c +13860a*b*c2+7920a*b*c*+495a*c*. Ex. 4. Required the term involving a1 in the expansion of (ax-bx3+cx3-&c.)1o. Here (ax-bx+cx3-&c.)1o=x1o(a−bx2+cx1-&c.)1o; therefore it will only be necessary to find the term involving in (a−bx2+cx1-&c.)1o. Hence the equations of condition are and by virtue of the second equation all the quantities after r must separately = 0. Ex. 5. Required the coefficient of x in (a + bx+cx2+dx3)3. .. coefficient required='(-1) (1-2)^ 13a + ( − 1 )a-4bc+ a¬3d, 8 EVOLUTION OF SURDS. A practical method of finding the square root of a binomial surd was given in Art. 182; the following is the one more usually adopted: 329. To extract the square root of a quantity which is under the form a+√b. Assume √x+√y=√a+√b, then squaring, x+y+2√xy=a+√b; .. x+y=a, and 2√ay=√,} (Art. 179): from these two equations we find x and y thus: x2+2xy + y2=a',\ .. x2—2xy+y2=a3—b, x-y=√a2-b. And x+y=a; .. 2x=a+√a2-b, |