350. Law of the Exponents.-The exponents of x and of a in the different powers follow a simple law. In the first term of each power, x is raised to the required power of the binomial; and in the following terms the exponents of a continually decrease by unity to zero, while the exponents of a increase by unity from zero up to the required power of the binomial. 351. Law of the Coefficients.—The coefficient of the first term is unity; that of the second term is the exponent of the power; and the coefficients of terms equidistant from the extremes are equal to each other; but after the first two terms it is not obvious how to obtain the coefficients of the fourth and higher powers. In order to discover the law of the coefficients, we will form the product of several binomial factors whose second terms are all different; thus, (x+a)(x+b)=x2+a|x+ab. +b (x+a)(x+b)(x+c)=x3+a | x2+ab | x+abc. to +ac (x+a)(x+b)(x+c)(x+d)=x2+a | x3+ab | x2+abc | x+abcd. +b +ac +abd In each of these products the exponent of x in the first term is equal to the number of binomial factors, and in the follow. ing terms continually decreases by one. The coefficient of the first term is unity; the coefficient of the second term is the sum of the second terms of the binomial factors; the coefficient of the third term is the sum of all their products taken two and two, and so on. The last term is the product of the second terms of the binomial factors. 352. We will now prove that if the laws of formation just stated are true for any power, they will also hold true for the formation of the next higher power. 2 Suppose that we have found the product of m binomials x+a, x+b,....x+k. Let P, denote the sum of the second terms of the binomials, P2 the sum of the different products of these second terms taken two and two, P, the sum of their products taken three and three, and so on; and let Pm denote the product of all these second terms. The product of the given binomials will then be -1. -2 3xm-3 Multiplying this polynomial by a new binomial, x+l, we ob tain the following product: The law of the exponents of x remains the same. The coefficient of the first term is still equal to unity, and that of the second term is the sum of the second terms of the m+1 binomials. The coefficient of the third term consists of the sum of the products of the second terms of the m binomial factors taken two and two, increased by the sum of the same second terms multiplied by 7, which is equivalent to the sum of the products of the second terms of the m+1 binomials taken two and two. The coefficient of the fourth term consists of the sum of the products of the second terms of the m factors of the first product taken three and three, increased by the sum of the products of their second terms taken two and two multiplied by l, which is equivalent to the sum of the products of the second terms of the m+1 binomials taken three and three, and so on. The last term is equal to the product of the second terms of the m binomial factors multiplied by l, which is equivalent to the product of the second terms of the m+1 binomials. Hence the law which was supposed true for m factors is true for m+1 factors; and therefore, since it has been verified for two factors, it is true for three; being true for three factors, it is also true for four, and so on; therefore the law is general. 353. Powers of a Binomial.-If now, in the preceding binomial factors, we suppose the second terms to be all equal to a, the product of these binomials will become the mth power of x+a. The coefficient of the second term of the product becomes equal to a multiplied by the number of factors; that is, it is equal to ma. The coefficient of the third term reduces to a2 repeated as many times as there are different combinations of m letters m(m—1) a2. taken two and two; that is, to 1.2 The coefficient of the fourth term reduces to a3 repeated as many times as there are different combinations of m letters m(m—1)(m—2) a3, and so on. taken three and three; that is The last term will be am. 1.2.3 Hence the mth power of x+a may be expressed as follows: m(m—1) a2xm−2+ 354. We perceive that if the coefficient of any term be multiplied by the exponent of x in that term, and the product be divided by the exponent of a in that term increased by unity, it will give the coefficient of the succeeding term. Forming thus the seventh power of x+a, we obtain (x+a)2=x2+7ax®+21a2x2+35a3x*+35a*x3+21a3x2+7aox+a2. We have thus deduced Sir Isaac Newton's Binomial Theorem. 355. In any power of a binomial x+a, the exponent of x begins in the first term with the exponent of the power, and in the following terms continually decreases by one. The exponent of a com mences with one in the second term of the increases by one. power, and continually The coefficient of the first term is one, that of the second is the ex ponent of the power; and if the coefficient of any term be multiplied by the exponent of x in that term, and divided by the exponent of a increased by one, it will give the coefficient of the succeeding term. 356. The coefficient of the nth term from the beginning is equal to the coefficient of the nth term from the end. If we change the places of x and a, we shall have, by the law of formation, The second member of this equation is the same as the second member of the equation in Art. 353, but taken in a reverse order. Comparing the two, we see that the coefficient of the second term from the beginning is equal to the coefficient of the second term from the end; the coefficient of the third from the beginning is equal to that of the third from the end, and so on. Hence, in forming any power of a binomial, it is only necessary to compute the coefficients for half the terms; we then repeat the same numbers in a reverse order. 357. The mth power of x+a contains m+1 terms. This appears from the law of formation of the powers of a binomial developed in Art. 352. Thus the fourth power of x+a contains five terms; the sixth power contains seven terms, etc. 358. The sum of the coefficients of the terms in the nth power of x+a is equal to the nth power of 2. For, suppose x=1 and a=1, then each term of the formula without the coefficients reduces to unity, and the sum of the terms is simply the sum of the coefficients. In this case (x+a)m becomes (1+1)m, or 2m. Thus the coefficients of the 359. To obtain the development of (x − a)m, it is sufficient to change a into a in the development of (x+a)m. In consequence of this substitution, the terms which contain the odd powers of a will have the minus sign, while the signs of the re maining terms will be unchanged. We shall therefore have m(m—1) a2xm-3__ (x—a)m—xm — maxm−1+ 1.2 Prefixing the coefficients, we obtain (a+b)ε=a+ba3b+15aab2+20a3b3+15a2b1+6ab3+b®. 2. Find the ninth power of a-b. The terms without the coefficients are ao, aob, a2b2, ao13, a5b1, a1b3, a3lo, a2b2, ab3, bo. The coefficients are 9x8 36x7 84x6 126x5 126x4 84x3 36×2 9x1 1,9, that is, 841, 36, 9, 1. 1, 9, 36, 84, 126, 126, Prefixing the coefficients, we obtain (a—b)3=a3 — 9aob+36a1l2—84ab3+126a5b1 — 126a1l3+84a3b® -36a2b79ab8-b9. It should be remembered that it is only necessary to com pute the coefficients of half the terms independently. 3. Find the seventh power of a—x. 4. Find the third term of (a+b)13. |