for the required power. This may be put under the form a3+b3+3ab(a+b). 88. Cor. 1. Hence, a quantity may be raised to any power, by multiplying its index by the index of the power; that is, (am)" amn. For the continued product of am into itself ʼn times, is found by adding the index m (Art. 35) as often as there are units in n; that is, multiplying m by n. 89. Cor. 2. Hence, also, any power of a product is equal to the product of the same powers of its factors, that is, (ab)" = a"b"; for (ab)" = ab × ab × ab &c. to n terms axaxa &c. to n terms xbxbxb &c. to n terms (Art. 11) = a"b". 90. Cor. 3. Hence also it follows that, when the quantity to be involved is negative, all the even powers will be positive, and all the odd powers negative. 91. According to the preceding rule, the successive powers of the binomial a+b, found by continual multiplication, will be as follows: (a+b)2=a2+2ab+b2. (a+b)3 = a3+3a2b+3ab2+b3. (a+b)1 = a1+4a3b+6a2b2+4ab3+b1. (a+b)5a5+5a4b+10a3b2+10a2b3+5ab1+b3. (a+b)6a6+6a5b+15a+b2+20a3b3+15a2b1+6ab5+b6. (a+b)=a7+7a6b+21a5b2+35a4b3+35a3b1+21a2b5+ 7ab6+b7. By comparing these developments it appears that, 1. The index of a decreases, and that of b increases, by a unit in each successive term; so that the sum of the indices in each term is always equal to the index of the given power. Hence, in general, the terms of (a+b)", without the coefficients, will be a", a"-1b, a"-2b2, a"-3b3, a"-4b4, a"-5b5, &c. .. ab"-1, b".* 2. The number of terms is one greater than the index; for there must be as many powers of a as there are units in the index, and one term besides, which does not contain a; so that when the index is even, the number of terms is odd, and vice versa. 3. The coefficients increase as far as the middle term, and then diminish, the terms equidistant from it having the same coefficient: thus those of the first and last are unity, of the second and last but one, the same as the given index, &c. And in general, if we know the coeffi cient in any term, that of the next following will be found by multiplying this coefficient by the index of a in that term, and dividing the product by the number of terms. Thus, in the case of (a+b), the coefficient of the second term is 6, the index of a in that term is 5, the efficient of the fourth term; and on examining the coefficients of the other powers, we shall find that they may be formed in the same manner. all According to this law, the coefficients of the terms of (a+b)" will be as follows: * The first and last terms may (Art. 84), be written abo, and The coefficient of the third term The coefficient of the fourth term n(n-1) n-2 2 3 -3 n(n−1) (n—2) (n—3) Combining these with the terms of (a+b)", already This is the celebrated BINOMIAL THEOREM, by means of which a binomial quantity may be involved to any power without the trouble of continual multiplication.* 92. If the second term, b, of the binomial be negative, its even powers will be positive, and its odd powers negative (Art. 90); but the even terms of the expanded power contain the odd powers of b, and therefore the powers of a being all positive, the odd terms of the powers of a-b will be positive, and the even terms negative. * A general demonstration of the binomial theorem will be given in a treatise on the principles of the higher Algebra, which will be published as a sequel to this work. It is convenient to introduce the theorem here for practical purposes; and the above inductive proof is the less objectionable, as no subsequent demonstration is founded upon it, and as it seems to have been by a similar process that Newton himself inferred the generality of this important formula. L In other respects, the powers of (a—b)" will be the same as those of (a+b)". Further, it is plain from the manner in which the developments are generated, that the series will terminate at the (n+1)th term, when n is a whole positive number, since, on passing that term, n-n or zero would always occur as a factor, and therefore every higher term would vanish. But assuming that the series holds when n is negative or fractional, n will not, in this case, be destroyed by any of the numbers successively taken from it. Thus no factor of the coefficients can become zero, and therefore the series will never terminate. EXAMPLES. 1. Required the 6th power of 2x2—3y. In this case we must put 2x2,-3y, and 6, for a, b, and n, in the general formula. Hence, The first term is The second ... The third The fourth .... The fifth The sixth .. ...... The seventh 6x (2x2)5(-3y)=-576x1oy. 6XX(2x2)X(-3y)2=2160x3y2. ... 15ק×(2x2)3 × (—3y)3——4320x6y3. 20 × 3 × (2x2)2 × (—3y)1—4860x*y*. 15x2x2x-3y)5-2916x2 y3. .................... 6 × † × (2x2)° × (—3y)6=729y6. Combining these, we have (2x2 — 3y)6 = 64x12 —576x1oy+2160x3y2 —4320x©y3+ 4860x1y1—2916x2y3+729y6. Xx 2. It is required to expand (a+b)2 by the binomial pand (a+b)-2, and then multiply all the terms by . That the series will never terminate may also be shown by actual division; for (Art. 81) (a+b)−2, or may be expanded into an infinite series. We have therefore 1 (a+b)2. (a+b)-2a22a3b+3a-4b2 — 4 a ̄5 b3 +5 a ̄6 b1 &c. x by a2+2ab+b2 (Art. 81). 3. Raise a+b+c to the third power. Regarding a+b as a single term, we may express (a+b+c)3 under the form {(a+b)+c}3; then |