Binomial Theorem. If the coefficient of any term is multiplied by the exponent which the first term of the binomial has in that term, and divided by the place of the term, the result is the coefficient of the next following term. 135. Corollary. If x is changed into - x in the preceding formula, it becomes 136. Corollary. The preceding formula written in the reverse order of its terms must give (x+a)" = x2+naxn-1+ whence it appears that n(n-1) 2 a2 xn - 2 + &c. The coefficients of two terms which are equally distant, the one from the first term, and the other from the last term, must be equal. Binomial Theorem. But, by the above formula, 2 (x - y) = x6 - 6 x5 y + 15 x4 y2 - 20 x3 y3 + 15 xo y4 -6xy5+y6; in which, if we substitute the values of x and y, we have (2ab2c-4b2cd)6 = 64 a6 b-12 c6 - 48 a5 b-9c7d+ 15 a4 b6 c& d2 — § a3 b-3 co d3 + a2 c10 d4 - 물이 a b3 c11 d5 + 시이이어 66 c12 d6. 2. Find the 10th power of a + b. Ans. a 10+ 10 a b + 45 a b + 120 a7 63 + 210 a6 b4 + 252 a5 b5 + 210 a4 b6 + 120 a3 b2+45 a2b+10 a 69+610. 3. Find the 11th power of 1 Ans. 1 - 11 x + 55 x2 x. 165 x3 + 330 x4 - 462x5+ 330 x7 + 165 x8 — 55 x9 + 11 x10 — 211. 4. Find the 4th power of 5 - 4 х. Ans. 625-2000 x + 2400 x2 - 1280 x3 + 256 x4. 5. Find the 7th power of 1 x + 2y. Ans. 2x + zx y + 2 x 5 y2 +35 x4 y3 +70 x3 y +168 x2 y5 + 224 x y + 128 y7. 6. Find the 4th power of 5 a2 c2 d - 4abd2. Ans. 625 a8 c& d4 2000 a b c d 5 + 2400 a6 b2 c4 d6 - 1280 a 5 63 c2 d7 + 256 a4 64 d8. 137. Problem. To find any power of a polynomial. Solution. Suppose the terms of the given polynomial to be arranged according to the powers of any letter, as x, as follows; a + a'x + a" x2 + a" x3 + av x4 + &c. Polynomial Theorem. in which the coefficient of any power of x is represented by the letter a, with a number of accents denoted by the exponent of this power. Suppose now the required power to be the nth, and if we denote the aggregate of all the terms of the polynomial except the first by X, we have and X = a x + a" x2 + a'' x3 + av x4 + &c.; (a + a' x + a' x2 + &c.) n = (a + X)". Now, by the preceding article, (a+x) = a2+nan-1 X + n (n-1) 2 an-2 X2 + n(n-1)(n-2) an-313+ n (n-1)(n-2) (n-3) an-4X4 + 2 3 2 3 4 n(n-1) (n-2) (n-3) (n-4) an-5-15 + &c. 2 3 4 5 We may now, by simple multiplication, obtain some of the first terms of X2, X3, X4, &c. as follows : X = a'x + a" x2 + a'' x2 + av x2 + av x5 + &c. a'x + a" x2 + a" x3 + a' x4 + av x5 + &c. X2 = a/2 x2+2a' a" x3 + +2a'a'''/x4+2a'a x5 +&c. + a" 2 2+2a α' x + α' x2 + a''' x 3 + av x4 + &c. X3 = a3x3+3a2a" x2+3a2 a'" \x5 + &c. Polynomial Theorem. in which terms containing powers of x higher than the 5th have not been retained. By substituting these values in (a + X)", and arranging the terms according to the powers of x, we obtain the 6 first terms of the required power as follows: (a+a'x+a" x2 + a' x3 + a x + av x5 + &c.)n If the coefficients of the different powers of x are compared with each other, we shall find that each can be obtained from the preceding term by the following process. Polynomial Theorem. From a given coefficient to obtain the next following coefficient : Multiply each term of the given coefficient by the exponent of the highest accented letter contained in this term, diminish the exponent of this letter by unity, and then multiply the term by that letter of the given polynomial which has one accent more than this which is at present the highest accented letter of the term. Again, whenever the accents of those two letters of a term which have the greatest number of accents differ in number by unity, that is, whenever the two highest accented letters are consecutive, multiply the term by the exponent of that one of these two letters which has the least number of accents, and diminish its exponent by unity; and at the same time increase the exponent of the other of these two letters by unity, and divide the term by the exponent thus increased. Thus the term Ta[e] would, if the accent e were higher than that of any other letter in this term, furnish by the first part of this rule p.Ta [e] p-1 a[+1] for the corresponding term of the succeeding coefficient. Also the term Ta [e] ra [+1]q would furnish the two terms and q. Ta[e] ra[+1] q-1 a[+2] P Ta[e] r-1 a[e+1] 1+1. By this method of deriving each coefficient from the preceding one, any power of a polynomial can be written down with great rapidity, and the correctness of this method can be demonstrated independently of the inductive process. |