the last be; in the cube, the first term is a3, and the last b3; and so on of the rest. II. That, with respect to the intermediate terms, the powers of a decrease, and the powers of b increase, by unity in each successive term. Thus, in the fifth power, we have In the second term, third, fourth, fifth, and so on in other powers. III. That in each case, the coefficient of the second term is the same with the index of the given power. Thus, in the square, it is 2; in the cube, it is 3; in the fourth power, it is 4; and so on of the rest. IV. That if the coefficient of a in any term be multiplied by its index, and the product divided by the number of terms to that place, this quotient will give the coefficient of the next term. Thus, in the fifth power, the coefficient of a in the second term multiplied by its index, and divided by the number of terms to that place= 4 X 5 = 20 2 = 10= coefficient of In the sixth power," Coeff. of a in the 4th term Xits index number of terms to that place. the third term. 20 × 3 60 4 4 15 coefficient of the fifth term. Hence, we are furnished with the following general rule for raising a binomial or residual quantity to any power, without the process of actual multiplication. RULE II. 287. Find the "terms without the coefficients, by observing that the index of the first, or leading quantity, begins with that of the given power, and decreases continually by 1, in every term to the last; and that, in the following quantity, its indices are 1. 2, 3, &c. Then, find the coefficients, by observing that those of the first and last terms are always 1 ; and that the coefficient of the second term is the index of the power of the first; and, for the rest, if the coefficient of any term be multiplied by the index of the leading quantity in it, and the product be divided by the number of terms to that place, it will give the coefficient of the term next following. Ex. 1. Required the 3th power of a+b. Here the terms, without the coefficients, are a3, a ̃b, aob2, a3b3, aaba, a3b3‚ ̄a2bo, ab1‚ b3. And the coefficients, according to the rule, will be And thus we have, (a+b)=a®+8a*b+28a b2 + 56a5 b3+70a3b1+56a3b3 +28a2b® +8ab2 +b*. 288. From this example and the foregoing Table the whole number of terms will evidently be one more than the index of the given power; after having calculated therefore as many terms as there are units in the index, of the given power, we may immediately proceed to the last term. And in like manner it may be observed, that when the number of terms in the resulting quantity is even, the coefficients of the two middle terms is the same; and that in all cases the coefficients increase as far as the middle term, and then decrease precisely in the same manner until we come to the last term. By attending to this law of the coefficients, it will only be necessary to calculate them as far as the middle term, and then set down the rest in an inverted order. Thus, in the above example, the middle term is 70a4b, and we have, The first four coefficients, 1, 8, 28, 56. 289. But we are not yet arrived at the most general form in which this Rule may be exhibited. Suppose it was required to raise the binomial a+b to any power devoted by the number (n). Proceeding with n as we have done with the several indices in the ceding examples, it appears that, The first term would be a". The second, pre The third, The fourth, The fifth ... 2X3X4 The sixth, n(n-1)x(n-2), 2X3 n(n-1)× (n-2) × (n—3) n(n-1)x(n-2) × (n−3) × (n—4)-575. 2X3X4X5 na -1b. n(n-1) Lan-26. 2 n(n−1) × (n−2) an−3f3+n(n−1) × (n−2) × (ñ—3) 2.3 an--4b4&c. 2.3.4 By the same process, (a-b)"=a”—na"-ib÷ n(n-1)x(n-2) × (n−3) 2.3.3. n(n-1)x(n-2), 2.3 an-4b4-&c.; the signs of the terms being alternately + and -1; and the sign of the last term is + or -1, according as n is even or odd; we have the last term in the former case jś +b", and in the latter —b". This general and compendious method of raising a binomial quantity to any given power, is called from the name of its celebrated inventor, Sir Isaac Newton's "Binomial Theorem." The demonstration of this Theorem, with its application to the finding the powers and roots of compound quantities, forms the subject of another Chapter. Its present use will appear from the following Example. Ex. 2. Required the fifth power of x2+3y2. Substituting these quantities for a, b, n, in the foregoing general formula, it appears, that The first (a") term, 2nd, } (nan-1b). is 5X (x2)1× 3y2 2 • =15x3y2. 4 90xby'. (n(n-1) × (n-2) an-3). is 5×2×3× (x2)2 5th, . (n(n−1)(n—2)(n—3) at—^ b 4 Last, (b′′) ・・・ is (3y2)3 ♦ So that (a2+3y2)5=x1°+15x3y2+90x y +-270x y+405x2y+243y10. 290. By means of this Theorem, we are enabled to raise a trinomial, or quadrinomial quantity to any power, without the process of actual multiplication. Ex. 3. Required the square of a+b+c? Here, including a+b in a parenthesis (a+b), and considering it as one quantity, we should have (a+b +c)2= [(a+b)+c]; and comparing them with the general formula; we have, (a")=(a+b)2=a2+2ab+b2 (nan1b)=2(a+b)×c=2ac+2bc (b")=c2 Hence, (a+b+c)2=(a+b)2+2(a+b)×c+c2=a2 +2ab+b2+2ac+2bc+c2. Ex. 4. Required the seventh power of a-b. Ans. a7-7ab+21a5b2-35ab3+35a3b4-21a2b +7ab°—¿1. 6 Ex. 5. Required the sixth power of 3x+2y. Ans. 729x+2916x5y+4860x1y2+4320x3y3 +2160 x2y1 +576xy5+64y6. Ex. 6. Required the square of x+y+3z. Ans. x+2xy+y2+6xz+6yz+9z2. Ex. 7. Required the fifth power of 1+2x. Ans. 1+10x+40x2 +80x3 +80x1+32x5. Ex. 8. Required the cube of x2-2xy + y2. Ans. x-6x5y+15x+y3 —20x3y3 +15x2y1 —6x¥3 +yo. § EVOLUTION OF ALGEBRAIC QUANTITIES. 291. The quantity which has been raised to any power is called the root of that power; thus the mth root of a power, is that quantity which we must continually multiply into itself, till the number of factors be equal to m m being a positive whole number, in order to produce the power proposed. We may conclude from this definition, and from the Articles in the preceding section. |