n Therefore, when x=y, we have nnxn-15-n--, 1-Y 1-Y or restoring m, mm P 9 P e 2cm Then, XX 9 1 1 Let 27 =v, and y?=u; then substituting we have =, ator and denominator by v—u.) But p and q are positive integers; therefore, when x=y and consequently v=u, we have THIRD. Let m be a negative fraction, or m= P р. 9 Let 29 =v, and y?=u; then a 3=, =0, and y=u, (Art. 8,) which values substituted give numerator and denominator by v—u.) But is a negative inte ger, and q is a positive integer; therefore, by what has already been shown, when x=y and consequently v=u, we have 33. It is required to obtain a general formula expressing the value of (a+b)", whether m be integral, fractional, positive or negative. In other words, it is required to obtain the general development of (a+b)m. In order to simplify, we resolve the binomial into two factors, thus; b b =am 1+ . ) (a+b)-= (1+)", we have only to 6 If then we obtain the development of (1+ multiply it by am to obtain that of (a+b)m. a 6 Let =x, then to develop (1+x)m assume (1+x)=A+BX+CX®+Dx3+ &c., (1) in which A, B, C, D, &c. are indeterminate coefficients, independent of x, and we are now to determine their values. * Now this equation must be true for any value of X; therefore, let X=0, and we have (1)m=A, or A=1. Substituting this value in (1), we have (1+x)"=1+BX+Cxo+Dx® + &c. (2) Again, since the development is to have the same form for all values of x, let x=y; then (2) becomes (1+y)=1+By+Cy' + Dyo+ &c. (3) Subtract (3) from (2), (1+2)/(1+y)m=B(x-y)+C(22—)+D(x3—48) + &c. Divide this by X-Y, (1+x)–(1+y)"=B+C. **—yo. + &c. (4) XY X-Y Let 1+x=v, 1+y=u; then x-y=v—u, which values substituted in the left hand member of (4), give co-go +D. (5) X-Y X-Y But when x=y, v=u, and we have, by Art. 32, whatever be the value of m, positive, negative, integral or fractional, +0. vym -um X?—Y V-u And we have also for the quotients in the right hand member of (5), when x=y, by Art. 32, x2 - y2 x x 23—Y: =2x2–1=2x, =323-1=3:2%, &c. xy X-2 * The coefficients A, B, C, D, &c., being indefinite, the expression A+ Bx + Coc2 + &c., may be assumed to be equal to any quantity whatever. It is then to be shown what values these coefficients must have, in order to satisfy the equation. These values substituted in (5), give m(1+x)=B+2Cx+3D:x° +4Ex + &c. (6) Multiplying this equation by 1+x, we have m(1+x)=B+2Cx+3DxP+4Ex3+ &c. + Bx +2Cx? +30x3+ &c. =B+(2C+B)x+(3D+2C)x2+(4E+3D)x3+ &c. (7) But if we multiply (3) by m, we have m(1+x)"=m+mBx+mCx+mDx + &c. (8) Forming an equation from (7) and (8), we obtain B+(2C+B)x+(3D+2C).x2 +4E+3D).x+ &c.= m+mBx+mCx2 +mDix3+ &c. (9) But this equation is of the form treated of in Art. 28; for its coefficients are quantities independent of X, and the equation is true for all values of X. Therefore, the coefficients of the like powers of X are equal each to each, and we have B=m whence B mB-B B(m-1) m(m-1) 2C+B=mB 66 C= 2 mC-2C_C(m-2)_m(m-1)(m-2) 3D+2C=mc - D= 3 2 3 mD-3D_D(m-3)_m(m–1)(m-2)(m-3) 4E+3D=MD 66 E= 2 3.4 &c. &c. These values of B, C, D, E, &c. substituted in (2), give m(m—1) m(m-1)(m-2) (1+x)=l+mx+ 2 2 3 m(m-1)(m-2m-3) -x++ &c. (10) 2 3 4 which is the general development of (1+x)". The law of the series is evident, and any number of terms may be written at pleasure. It only remains to find the development of (a+b)m. In (10) restore b the value of x= we have m 2 a ( b m(m—1) 62 , m(m-1)(M—2) b3 + 2 . 3 it a +&c. 2 Finally, multiplying by a", we obtain (m(m—1) (11) 2. 3 which is the general formula for the development of any binomial (a+b)", whatever be the values of a and b, and whether m be positive, negative, integral or fractional; and this formula is known as the Binomial Theorem of Sir Isaac Newton. 34. If in formula (10) x be made negative, the uneven powers of X will be negative and the even ones positive, (Art. 14), so that the formula will become m(m-1) m(m-1)(m-2) (1-x)=1-mx+ 28+ &c. (12) 2 2 . 3 Formulæ (10) and (12) may be expressed in one formula, thus ; m(m-1) (1+x)=l=mx+ m(m-1)(-2) X? + 23 + &c. (13) 2 2 3 22 35. If we make b negative in (11) we shall find in the same manner (a—b)m=am-mam–16+ -1) (14) 2 Expressing (11) and (14) in the same formula, (a+b)m=amŁmam–18+m(m—1) am_252+ &c. 2 Formula (13) is only a particular case of (15) when a=1 and b=x. (15) 36. In assuming our series we included all the powers of x, from O upwards; so that the above developments are series of an infinite number of terms. But when m is a positive integer, the series will terminate at the (m+1)th term, and all the succeeding terms will vanish, or become =0. For by examining formula (15), it will be seen that the 2d term contains the factor m, the 3d term the factor m—1, the 4th term the factor m—2, &c., and the (in+2)nd term will contain the factor in-m or 0, which reduces that term to 0; |