Take r, therefore, the first whole number greater than (n+1); and the 7th term will be the greatest. b 'a+b' n-r+1 b a a+b be a whole number, when r is equal to it we shall have -1, and then two terms, the 7th and the (r+1)th will be equal, and each of them greater than any of the other terms. If the index be negative, n, the rth term will be the greatest when -n-r+1 b n+r-1 b is first <1, irrespectively of sign; i. e. when is first a And if this is a whole number, two terms will be equal, and each greater than any of the others. COR. By thus ascertaining the greatest term we determine the point from which the terms of a series become less and less, or, as it is usually stated, the point at which the series begins to converge. Ex. Required to find which is the greatest term in the expansion of (3+5x), when x= 1 2 The first whole number greater than 4 is 5; therefore the term required is the 5th. 320. To find the sum of all the coefficients of an expanded binomial. Since (1+x)^=1+nx+”("~1) '+&c., for any value whatever of x, let 2 (1+1)"=1+n+ " (n−1) + n (n−1) (n−2) + &c. 1.2 1 2. 3 or 2" the sum of the coefficients. Ex. (x+a)=x+5ax1+10a2x3+10a3x2+5a*x+a, and the sum of the coefficients=1+5+10+10+5+1=32=23. Also by putting x=-1, we have or 0=sum of the odd coefficients - the sum of the even ones. .. the sums of the odd and even coefficients are equal, and consequently each2" 2" rem. take 321. To find the approximate roots of numbers by the Binomial Theo The theorem being proved for a fractional index, we have Now, if N represent a proposed number whose nth root is required, P such that p" is the nearest perfect nth power to N, so that N=p" q, q being small compared with p", and + or - according as N> or <p"; of which series a few terms only will give the required root to a considerable degree of accuracy. Ex. Required the approximate cube root of 128. 322. A trinomial, a+b+c, may be raised to any power by considering two terms as one, and making use of the Binomial Theorem. Thus, (a+b+c)" =(a+b+c)", in which the several powers of a+b may be replaced by their expansions found by the Binomial Theorem. Ex. Required the cube of 1+x+x2. (1+x+x3)3=(1+x+x2)3, =(1+x)3+3(1+x)2x2+3 (1+x) x*+x®, =1+3x+3x2+x3+ 3x2+6x3+ 3x2+3x′+ 3x2+x®, =1+3x+6x+7x+6x*+3x+x®. Similarly (a+b+c+d)" may be expanded by considering a+b as one term and c+d as another; and any multinomial may be expanded in a similar manner by dividing the whole into two terms and considering it as a binomial. Also any particular term of an expanded trinomial may be easily found:-thus To find the term involving a' in the expansion of (1+x+x2)3. (1+x+x2)3=(1+x)3 +3(1+x)2x2+3(1+x)x*+x®, and without further expansion it is seen that the term required = 3x2× x2+ 3x1=6x*. Again, the number of terms in the expansion of (a+b+c)" may be found; for it will obviously be the aggregate number of the terms in the expansions of the several powers of a+b, from (a+b)" down to (a+b)o, that is, (Art. 317) =(n+1)+n+(n-1)+ &c. to n+1 terms, = {2n+2−n}”+1 (Art. 282) = [Exercises Zg.] (n+1)(n+2) 1.2 323. Required to find the Remainder after taking r terms of the expansion of (1−x)→. By the Binomial Theorem, (1-x)=1+2x+3x2+...+rx2-1+R, R representing the remainder after r terms; In the same manner may be found the remainder after taking r terms of the expansions of (1-x)-3, (1−x)^, &c. 324. To find the number of homogeneous products of r dimensions which can be made of n things a, b, c, d, &c. and their powers. By common division, or the Binomial Theorem, the coefficient of a being the sum of the homogeneous products of n things a, b, c, d, &c. of r dimensions. Now to obtain the number of these products, let a=b=c=d=&c.=1, then the coefficient of a will give the number required. But on this supposition, the left-hand side of the equation becomes (1-x)", which by the Binomial Theorem is COR. Hence also the number of terms in the expansion of any multinomial, as (a,+a2+a ̧+.....a‚)", is known; for it is obviously the same as the number of homogeneous products of r things taken n together, that is, If r=2, that is for a binomial, (a+b)", the expression becomes If r=4, or the quantity to be expanded be (a+b+c+d)", the number of terms is 4.5.6......(n+3) (n+1)(n+2)(n+3), 1.2.3... and so on for any value of r. n ; 1. 2 3 THE EXPONENTIAL THEOREM. 325. To expand a2 in a series of powers of x. a2={(1+a−1)"}"; and expanding by the Binomial Theorem, ➡{1 +n(a−1)+n. ” — 1 (a−1)'+n.”—1.^= 2 (a−1)*+&c.}", n-1 2 3 ={1 +[(a−1) − (a−1)2 + (a−1)° - &c.]n+Bn'+ Cn2+...}", B, C, &c. 2 containing powers of a-1 only; 3 1 ={1+An+Bn2+Cn3+&c.}", if a−1- x n = 1 + 2 (An+ Bn2+ ... ) + ~ . ~ 2 (An+Bn2+... . . )2 + ... ... ... ... .... n . . =1+x(A+Bn+.....)+x.~,” (A+Bn+........)'+................ 2 Now, since a is clearly independent of n, n must entirely disappear from the above series, which will therefore consist only of those terms in which n is not found; COR. Ife be that value of a which makes A equal to 1, then 1 1 1 6=1+=+ + +...=2·71828, (Art. 57, Ex. 1). 1 1.2 1.2.3 Another method of expanding a* will be found in Art. 342, Ex. 9. THE MULTINOMIAL THEOREM. 326. The Multinomial Theorem is a rule or formula for expanding any power of a quantity which consists of more than two terms. The expansion of a multinomial may frequently be effected by the Binomial Theorem, as is done for a trinomial in Art. 322; for (a+b+c+d+&c.)" may be expanded as a binomial by considering any number of terms as one term, and the remainder as another term. But a more general method is to find the general term, and to deduce the whole expansion from that term as follows: 327. To find the general term of the expansion of (a+b+c+d+&c.)TM. Let b+c+d+&c.=2, then (a+b+c+d+&c.)TM= (a+2)", of which the general term, expressed by the a+1th, is Again, if c+d+&c.=y, then z=(y+b), of which the general term, expressed by the q+1th, is |