(x+a)9=x+9x3a+36x7a2 + 84x6a3 +126x3a1+126x*a5+84x3a® +36x3a2 +9xas+a9. EXAMPLE II. In like manner, (x-a)10-10-10x°a+45.x a2-120x7a3+210x a*—252x*a*+210x1a-120 xa+45x a-10.ra+a10. 108. The labor of determining the coefficients may be much abridged by attending to the following additional considerations: V. The number of terms in the expanded binomial is always greater by unity than the index of the binomial. Thus, the number of terms in (x+a)1 is 4+1, or 5; in (x+a)10 is 10+1, or 11. VI. Hence, when the exponent is an even number, the number of terms in the expansion will be odd, and it will be observed, on examining the examples already given, that after we pass the middle term the coefficients are repeated in a reverse order; thus, The coefficients of (x+a)1 are 1, 4, 6, 4, 1. The coefficients of (x+a) are 1, 6, 15, 20, 15, 6, 1. The coefficients of (x+a)s are 1, 8, 28, 56, 70, 56, 28, 8, 1. VII. When the exponent is an odd number, the number of terms in the expansion will be even, and there will be two middle terms, or two contiguous terms, each of which is equally distant from the corresponding extremities of the series; in this case the coefficient of the two middle terms is the same, and then the coefficients of the preceding terms are reproduced in a reverse order; thus, The coefficients of (x+a)3 are 1, 3, 3, 1. The coefficients of (x+a)5 are 1, 5, 10, 10, 5, 1. The coefficients of (x+a)7 are 1, 7, 21, 35, 35, 21, 7, 1. The coefficients of (x+a)9 are 1, 9, 36, 84, 126, 126, 84, 36, 9, 1. 109. If the terms of the given binomial be affected with coefficients or exponents, they must be raised to the required powers, according to the principles already established for the involution of monomials; thus, ..(2x+5a2)=16x12+160x9a2+600xa+1000x3a+625a3. In like manner, EXAMPLE IV.* (a3+3ab)=(a3)9 +9 (a3)8 × (3ab) +36 (a3)7 × (3ab) +84 (a3) X (3ab)3 +126 (a) (3ab)*+ 126(a3) × (3ab)5 +84 (a3)3× (3ab ) +36(a3)2×(3ab)7+9a3× (3ab)8+(3ab)9 =a27+27a2b+324ab2 +2268aa1 b3 +10206a19b4 + 30618a17b5 +61236ab6+78732a3b7+59049ab8+19683ab9. 110. We shall now proceed to exhibit the binomial theorem in a general form. Let it be required to raise any binomial (x+a) to the power represented by the general algebraic symbol n. Then, by the preceding principles, we shall have The whole number of terms will be n+1, and the coefficients be repeated in a reverse order after the (n+1)th, or (+1) th term, according as n is odd or even; moreover, the terms will all have the sign +, if the quantity to be expanded be of the form x+a, and they will have the sign + and nately, if the quantity be of the form x-a. Hence, generally, n(n-1) 1.2.3 'x2a"→2+nxa11+a" alter The best method of proceeding in these examples is to raise (y+z) to the fourth and ninth powers, and then, in the expansions thus obtained, to substitute 2x3 for y, and 5a2 for z in the first, and a for y, and 3ab for z in the second. In this last case, if n be an even number, the last term, being one of the odd terms, will have the sign+; and if n be an odd number, the last term, being one of the even terms, will have the sign Both forms may be included in one by employing the double sign. To exemplify the application of the theorem in this form, let it be required n(n-1)(n-2)(n—3) becomes =625c8 becomes 4(5c)3× (2yz) =1000c®yz 4.3 becomes 1.2(5c) X (2yz)2= 600c1y3z2 4.3.2 becomes 1.2.3 (5c2)1× (2yz)3= 160c2y3z3 4.3.2.1 a1 becomes: ¡(5c2)°× (2yz)1= 16y1z1 ... (5c-2yz)=625c-1000c yz+600c1y2z2-160c2y3z3+16y1z1. 111. We have sometimes occasion to employ a particular term in the expansion of a binomial, while the remainder of the series does not enter into our calculations. Our labor will, in a case like this, be much abridged, if we can at once determine the term sought, without reference either to those which precede, or to those which follow it. This object will be attained by finding what is called the general term of the series. If we examine the general formula, we shall soon perceive that a certain relation subsists between the coefficients and exponents of each term in the expanded binomial, and the place of the term in the series; thus, Observing the connection between the numerical quantities, it is manifest, that if we designate the place of any term by the general symbol p, the pth This is called the general term, because by giving to p the values 1, 2, 3, 4, ......... we can obtain in succession the different terms of the series for (x+a)". EXAMPLE VII. Required the 7th term of the expansion of (x+a)12. Here n=12}..n-p+2=7, n−p+1=6 P= 75 p-1=6. Substituting these values in the general expression, we find that the term sought is Since the second term of the proposed binomial has the sign all the even terms of the expansion will have the sign sign; therefore the 5th term is and all the odd terms the +1032192c20h20 EXAMPLE IX. Required the middle term of the expansion of (r—a)18. Since the exponent is 18, the whole number of terms will be 19, and hence * The operation here to be performed is best effected by canceling the factors. the middle term will be the 10th; and since it is an even term, it will have the sign ; hence it will be Required the third and the last terms of the expansion of (x+2y)1 TO EXTRACT THE nth ROOT OF A NUMBER. 112. The nth power of 10 is 1 with n ciphers, and the nth power of any number below 10 must be less, and can, therefore, be composed of no more than n figures. The nth power of 100 is 1 with 2n ciphers, and the nth power of any number between 10 and 100 can not, therefore, contain more than 2n figures, nor less than n. For a like reason, the nth power of three figures can not contain more than 3n, nor less than 2n. That of four figures can not contain more than 4n, nor less than 3n, &c. The nth root of a number being required, it is evident from the above that there will be as many figures in the root as there are periods of n figures in the given number, counting from right to left, and one more if any figures remain on the left. The root may be divided into units and tens, and the nth power of it, or the given number, will be equal, according to the Binomial Theorem, to the nth power of the tens plus n times the n-1 power of the tens into the units plus a number of other terms which need not be considered. Tens have one cipher on the right, and hence the nth power of tens has n ciphers on the right; then right-hand significant figures, therefore, make no part of the nth power of the tens; to find the tens of the root, then, the nth root of those figures which remain, after rejecting n on the right, must be sought by an independent operation; but if there are more than n of these remaining figures, the tens of the root are expressed by a number containing more than one figure, which number may be separated into its units and tens, the nth power of the tens of which does not contain the n significant figures on the right of that number upon which the independent operation is now performing, and in consequence these n figures are also rejected. After rejecting periods of n figures successively, beginning on the right until there remains but one period and part or the whole of another period on the left, let these be considered an independent number, its root will contain two figures, tens and units; the nth root of the tens is to be sought in what is left after rejecting the right-hand period; the n-1 power of the tens has n-1 ciphers on the right; so, also, has any multiple of this, and, therefore, n times the n-1 power of the tens into the units; which last quantity, therefore, is not to be sought in the n-1 right-hand significant figures; after subtracting the nth power of the tens just found, only one figure of the next period, therefore, is to be placed on the right of the remainder, which is then divided by n times the n-1 power of the tens; the quotient will not be exactly the units, for the dividend contains also a part of the other terms of the power of the binomial which were not considered; this quotient may be greater than the units of the root, but never can be less; it must be diminished till the nth power of the two figures found is equal to or less than |
