Binomial Theorem. Secondly. To find the form in which a enters into the development. Let a= 1, x = x', and let the corresponding values of a", B, C, &c., be 1, B', C', &c., and we have (1+x')"=1+ B' x' + C' x' 2 + D' x13 +&c. in which A', B', C', &c., must be independent of a. product of this equation by a" is The a" (1+x')"=(a+ax')"=a"+B' a" x' +С' a" x12+&c., in which, if we put we have and ax'x, or x'= a' a" x' = a^-1 x, a" x' 2 = an−2 x2, &c., -2 -3 (a+x)"=a"+B' aa¬1 x + C′ aa¬2x2+D'a1¬3 x3+&c. Thirdly. To find the coefficients. The derivative of the last equation is, by examples 4 and 8 of art. 175, n(a+x)n−1=B'an-1+2C'an-2x+3D'a1¬3x2+4E'a2¬1ñ3 &c., which, multiplied by (a+x), gives n(a+x)”=B'a”+2C"a"¬1 x+3D' a"−2x2+4E' a2¬3x3+&c +B'an 1x+2C'a¤¬2x2+3D' a"¬3x3+&c. The product of (a + x)" by n gives, also, n(a+x)" —na"+n B'an-1x+nC'a2¬2x2+n D'aˆ ̃3x3+&c. which, compared with the preceding equation, gives, by art. 163, B'=n, 2C+B'=nB', or 2 C'—(n−1)B', or C'={(n − 1)B' ; ́ ́ 3D'+2C=nC', 3D'=(n-2) C', _D'=}(n-2)C' ; 4E+3D nD', 4E' (n-3)D', &c. &c. E=(n-3)D'; &c. Binomial Theorem. The combination of these results gives Sir Isaac Newton's Binomial Theorem. The first term of any power of a binomial is the same power of the first term of the binomial. In the following terms of the power the exponent of the first term continually decreases by unity, whereas the exponent of the second term of the binomial, which is unity in the second term of the power, continually increases by unity. The coefficient of the second term of the power is the exponent of the power. 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. 212. Corollary. The equations of the preceding article give n (a + x)" = a” +n a"−1 x + " (n−1) ar−2x2+ 2 an-2x8+ n(n−1)(n—2) an−3.x3+ n(n−1)(n—2)(n—3) an-4x4+&c. 2 3 2. 3. 4 Binomial Theorem. 213. Corollary. If x is changed into ceding formula, it becomes (a — x)" — a” —n an−1 x + = n (n − 1) an-3x+ &c. the signs of every other term being reversed. 214. Corollary. The preceding formula, written in the reverse order of its terms, gives The coefficients of two terms which are equally distant, the one from the first term, and the other from the last term, are equal. 215. EXAMPLES. 1. Find the 6th power of 2ac — ‡ b c2 d. b2 (2 a c — + b co d) © = ( z − y)®. But, by the above formula, (x — y)6 — x6 — 6 x5 y + 15 x1 y2 — 20 x3 y3 + 15 x2 y^ -6x y5y; in which, if we substitute the values of x and y, we have Binomial Theorem. (2 ab-2c-4bc2 d)6 64 a6 b-12 c6-48 a5 b-9c7d+ 15 a4 b6 c8 d2 - § a3 b-3 c9 d3 + 15 a2 clo d c10 − 2 3 6 a b3 c11 d5 +4096 b6 c12 d6. 2. Find the 10th power of a+b. Ans. a10+10 a9 b+45 ao b2+120 a7 b3 +210 ao ba of 252 a5 b5+210 a1 b6+120 a3 b7+45 a2 b®+ 10 a bo + b10. 3. Find the 11th power of 1 x. Ans. 1—11x+55 x2-165 x3+330 x4-462x5+462 x6 -330x7165 x8-55 x911 x10x11. 4. Find the 4th power of 5-4 x. Ans. 625-2000 x2400 x2-1280 x3+256 x1. 5. Find the 7th power of x+2y. Ans. 12x+3z xo y + 21 x5 y2+ 3,5 x1 y3 + 70 x3 y1 +168 x2 y5+224 x yo + 128 y7. 6. Find the 4th power of 5 a2 c2 d 4 ab d2. · 2000 a7 b c d5 +2400 a6 b2 c4 do - 1280 a5 b3 c2 d2 +256 aa ba d3. 216. 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 + bx + c x2 + dx3 + ex2+&c., in which the successive coefficients are denoted by the suc cessive letters of the alphabet. The following is Arbogast's rule for finding any power of the poly nomial. Polynomial Theorem. The first term of the power is the same power of the first term a of the given polynomial. The coefficient of x in the second term is b times the derivative of the first term taken with reference to a. To obtain any other coefficient from the preceding coefficient; let r be the letter farthest advanced in the alphabet which is contained in any term of the given coefficient. Then r times the derivative of this term with reference to q, is a term of the new coefficient; but this process is obviously inapplicable, or at least useless, when q is the last letter of the given polynomial so that r is zero. If the given term contains the preceding letter p as well as q, q times its derivative with reference to p, divided by the increased exponent of q, is also a term of the new coefficient. Thus the term T p q gives, in the following coefficient, the two terms 0 T p q11r and — Tp-1 q+1. 6+L Proof. First. The value of the first term is obtained precisely as in the binomial theorem by putting x = 0. Secondly. Let V denote the given polynomial, so tha V= a + bx + c x2 ... +p x2 + q x*+1+&c. The derivative of V with reference to p is, then, x", and that of Vn is, by art. 172, n Vn-1 x*, |