108. The labour of determining the coefficients may be much abridged by at. tending to the following additional considerations · V. The number of terms in the expanded binomial is always greater by unity than the index of the binomial. Thus the number of terms in (x + a) * is 4+ 1, or 5, in (x + a)1o is 10 + 1, or 11. VL. Hence, when the exponent is an even number, the number of terms in the expansion will be odd, and it will be observed, on examining the examples already given, that after we pass the middle term the coefficients are repeated in a reverse order; thus, The coefficients of (1 + a)* are 1, 4, 6, 4, 1. (x+a) * 1, 6, 15, 20, 15, 6, 1. (a) — 1, 8, 28, 56, 70, 56, 28, 8, 1. VIL When the exponent is an odd number, the number of terms in the expansion will be even, and there will be two middle terms, or two contiguous terms, each of which is equally distant from the corresponding extremities of the series; in this case the coefficient of the two middle terms is the same, and then the coeficients of the preceding terms are reproduced in a reverse order; thus, The coefficients of (z+a) 1, 3, 3, 1. (x + a)1 (a) are 1, 5, 10, 10, 5, 1 1, 7, 21, 35, 35, 21, 7, 1. 1, 9, 36, 84, 126, 126, 84, 36, 9, 1. 109. If the terms of the given binomial be affected with coefficients or exponents, they must be raised to the required powers, according to the principles already established for the involution of monomials; thus: : ·22+5a2) = 16+160x'a2+600x*a* +1000x3a*+ 625a3 (e'+30% j' = (a2)' + Xa3)3 × (3ab)+36(a3)' × (3ah)2 +84 (a3 ) × (3ab)• +126(a)× (3ab) + 126(a')' × (3ab)3 +84(a3)3 ×(3ah +36 (a3 ja × (3ab)' + 9a3 × (3ab) + (3ab)9 = a+278b+324aTMb2+2268 aa1b3+10206a1b* + 30618a1b +61236 ab+78732 a112 + 59049 a11 b3 +19683 a'b' 1:0. We shall now proceed to exhibit the binomial theorem in a general form Let a be required to raise any binomial (r+a) to the power represented by BINOMIAL THEOREM. 105. It is manifest, from what has been said above, that algebraic polynomials may be raised to any power merely by applying the rules of multiplication. We can however in all cases obtain the desired result without having recourse to this operation, which would frequently prove exceedingly tedious. When a binomial quantity of the form x+a is raised to any power, the successive terms are found in all cases to bear a certain relation to each other. This law, when expressed generally in algebraic language, constitutes what is called the "Binomial Theorem." It was discovered by Sir Isaac Newton, who seems to have arrived at the general principle by examining the results of actual multiplication in a variety of particular cases, a method which we shall here pursue, and give a rigorous demonstration of the proposition in a subsequent article of this treatise. Let us form the successive powers of x+a by actual multiplication. 3 4 x+5x3u+10 x1 a2 + 10 x 3 a3 + 5 x2 a1 + xa3 5 3 3 2 4 x® + 6x 3a + 15 xa1 a2 + 20 x 3 a 3 + 15 x 2 a 1 + 6 x a 3 + a .........6th power. x + a x+6x6a+ 15 x3 a2+20 x1a3 + 15 x3 aa + 6 x2 a2 + x а + xo a+ 6 x3 a2 +15 x1 a3 + 20 x3 a* +15 x2 a3 4 +6 xa® + a1 x2 +7 x®a+21 x 3 a2 +35 x 1 a3 +35 x 3 a 1 +21 x 3 a3 +7 xa® +a2îth power. In order that these results may be more clearly exhibited to the eye, we shall arrange them in a table. (x+a)x+5x+a+10x3 a2+10x13+5xa+a3 (x+a) * x2 + 6 x3 a + 15 x1 a2 + 20 x3a3+15x2a1+ 6xa3+a' ̧(x+a) x2+7x® a +21 x *a* + 35 x1a3 +35 x3a*+21x2a3+7xa®+a" (z+a) x2+8x7a+28x® a3+56x3a3+70x*a*+56x3a3 +28x2a®+8xa1+a8 In the above table, the quantities in the left hand column are called the expressions for a binomial raised to the first, second, third, &c. power; the corresponding quantities in the right hand column are called the expansions, or, developements of the others. (x — a) 3 106. The developements of the successive powers of x a are precisely the same with those of x + a, with this difference, that the signs of the terms are alternately + and ; thus, 3 = x3 — 5 x 1 a + 10 x3 a2 10 x 2 a3 + 5 x a1 — a 3 and so for all the others. 107. On considering the above table we shall perceive, that I. In each case the first term of the expansion is the first term of the binomial raised to the given power, and the last term of the expansion is the second term of the binomial raised to the given power. Thus, in the expansion of (x + a) * the first term is 4 and the last term is a1, and so for all the rest. II. The quantity a does not enter into the first term of the expansion, but ap pears in the second term with the exponent unity. The powers of a decrease by unity, and the powers of a increase by unity in each successive term. Thus, in the expansion of (x + a) we have, x o, x 3 a, x1 a 2, x a x2 a 1, x a 3, a o. III. The coefficient of the first term is unity, and the coefficient of the second term is in every case the exponent of the power to which the binomial is to be 6 3 2 " raised. Thus the coefficient of the second term of (x + a)2 is 2, of (x + a) ® is 6, of (x + a)' is 7. IV. If the coefficient in any term be multiplied by the index of r in that term and divided by the number of terms up to the given place, the resulting quotient will be the coefficient of the succeeding term. Thus in the expansion of (x+a) • the coefficient of the second term is 4; this multiplied by 3, the index of x in that term, gives 12, which when divided by 2 the number of terms up to the given place gives 6, the coefficient of the third term. Again, 6 the coefficient of the third term multiplied by 2, the exponent of x in that term, gives 12, which, when divided by 3, the number of terms up to the given place, gives 4, the coefficient of the 4th term. So also 35, the coefficient of the 5th term in the expansion of (x+a), when multiplied by 3, the index of x in that term, gives 105, which, when divided by 5, the number of terms up to the given place, gives 21, the coefficient of the succeeding term. By attending to the above observations, we can always raise a binomial of the form (x+a) to any required power, without the process of actual multiplication, 7 (x + a)=x+9x a +36 x1 a2 + 84 x® a3 +126xa + 126 x 'a'+ x3 a +36 x 2 a2 + 9 x a 3 + a9. 8 In like manner, Example II. (x - a) 10 = 10 xa + 45 x ® ao 9 -252 x 5 a3 + 210 x1a' — 120 x3 a1 + 45 x2 a 8 — 10 x ao + « 1o. 108. The labour of determining the coefficients may be much abridged by at. tending to the following additional considerations. V. The number of terms in the expanded binomial is always greater by unity than the index of the binomial. Thus the number of terms in (x + a) is 4+ 1, or 5, in (x + a)1o is 10 + 1, or 11. VI. Hence, when the exponent is an even number, the number of terms in the expansion will be odd, and it will be observed, on examining the examples already given, that after we pass the middle term the coefficients are repeated in a reverse order; thus, The coefficients of (x + a) VII. When the exponent is an odd number, the number of terms in the expansion will be even, and there will be two middle terms, or two contiguous terms, each of which is equally distant from the corresponding extremities of the series; in this case the coefficient of the two middle terms is the same, and then the coefficients of the preceding terms are reproduced in a reverse order; thus, The coefficients of (x+a)3 are (x + a) 6 second third fourth are 1, 4, 6, 4, 1. fifth 1, 6, 15, 20, 15, 6, 1. 1, 8, 28, 56, 70, 56, 28, 8, 1. 1, 3, 3, 1. 1, 5, 10, 10, 5, 1 1, 7, 21, 35, 35, 21, 7, 1. 1, 9, 36, 84, 126, 126, 84, 36, 9, 1. 4. 4 (2x3)4 6 X 2 = · (2 x 3)1 × (5 a2)3 4X2 X 125 x3 aa. 3 (2 x 3) ° × (5 a 3) 4 16 x 12 = 4 X 8 X 5 x9a2 2 6X4 X 25xo aa. = = 625 a 8 16.12160x9a2 + 600xo a1 + 1000x3 a® + 625a3 6 Example IV. In like manner, (a3+3ab) = (a3)9 + 9(a3 )8 × (3ab)+36(a3)' × (3ab)2 +84 (a3 ) × (3ab) * +126(a) x (3ab) + 126(a)1 × (3ab) +84(a)3 X(3ah)® +36 (a3 )2 × (3ab)' + 9a3 × (3ab)3 + (3ab)9 = a+27a2b+324ab2+2268 63+10206a1b + 30618ab +61236 a1566 + 78732 a13b2 + 59049 a11 b8 +19683 a9b9 110. We shall now proceed to exhibit the binomial theorem in a general form Let it be required to raise any binomial (r+a) to the power represented by |