3. Find the 4th power of 3a%c—2bd. Ans. 81a8c4_216a6cbd+216a*cob?!?_96aocb%d+1664d4. 41. Expansion of binomials with negative integral exponents. This is effected in the same way as in the case of positive integral exponents, by substitution in formulæ (13) and (15), these formulæ having been proved true for any value of the exponent. Therefore, 1 to expand (a+b)-2, or (Art. 20,) at6 in (15) let m=-1, and we have (a+b)-=a-14a-4b40-352—a-463+ &c., + + &c. a+b a> a3 which is, by Art. 37, an infinite series. The same result will be obtained by actual division : 1 atb b 62 1+ + &c. or a b1 aa &c. from which it is also evident that the series will never terminate. 1 To convert the fraction into an infinite series; in other 1+x words, to expand (1+x)-1. In (13) let m=-1. Then we have (1+x)-1=1-2+x-x3+ &c. 1 =1—3+x2–23+ &c. 1+2 If in this equation we make x=1, it becomes 1 &c 2 or This singular series is only the limit of a large-class of series which are divergent, or whose terms instead of constantly decreasing go on increasing to infinity; so that the sum of any number of terms can never express the value of the series. In the above example the sum of 2 terms is 0, of 3 terms 1, of 4 terms 0, &c., from which nothing more can be inferred than that the value of the series is between 0 and 1.* 42. A series, however, is readily obtained which correctly ex 1 1 presses the value of the fraction Take the fraction 1 2 1 X+ and in (15) let a=x, b= 2 m=-1; we have 2 or X + + &c. 4 (x+)* ()+() 5(.)*+&c. or the rules of probabilities, take the mean, which gives 1 * The distinguished LEJBNitz, however, according to LaPlace, imagined 1 that the form of this series warranted the deduction from it of the value 2 He observed, “that the series becomes 0 or 1, according as we take an even or an uneven number of terms; and as in infinity there is no reason for preferring the even to the uneven number, we should, according to 1 for the value of 2 1+ the series.” But this reasoning is fallacious; for if the fraction 1++x2 be converted into a series, it will be found to be 1-23 + x3—205 + - &c., 2 and when x=1, the fraction becomes and the series 1-1+1-1+ &c. 3 It appears, therefore, that the value of this series is indefinite, and that it may correspond to various fractions between 0 and 1.-Laplace, Essai Philosophique sur les Probabilités. 81 by only 1458 -n-1) + &c. 2 2 2 2 + &c. 1 The sum of 4 terms of this series is which differs from 2 by 1 364 1 the 162 sum of 6 terms is which differs from 729' 2 and the greater the number of terms we add together, the nearer we 1 approximate to the value which is therefore the value of an in. 2 finite number of terms. A series of this kind, in which the terms decrease to infinity, is called a convergent series. 43. If in (13) we make m=-n, we shall find (1kg)-"=linc+ (-1)(-n-1) (-2)(-n-1)(-n-2) -23+ &c. 2 2.3 n(n+1) n(n+1)(n+2) (16) 2 2.3 If in (15) we make m=-n, we shall find in the same manner (a+b)-*=*=*=na---26 +"(n an n(n+1)(n+2) a-n-363+ &c. 2.3 or by Art. 20, 1 1 b , nbo + +"ran 2 c.) (17) 2.3 These two formulæ (16) and (17) may be used to expand binomials with negative integral exponents instead of the general formulæ ; and will be found much more convenient in practice. y 1 44. Thus, to expand or yx By (17), making (3x+y) (3x+y) a=3x, b=y, n=2, we find 1 1 y Y 2.3.4 Зх, 2.3 Зr, ข 2y,3y2 4y5y &c. + n(n+1) ----2697 azt ast &c. 3x+yje= + 45. Expansion of binomials with positive fractional exponents. It will be convenient here, as in the last case, to obtain a formula for fractional exponents. In (13) let m= , we find 22+ C-1) (1+x)ả=1+Le+ 2C-1)(-2) zř=1x \ 2x24P(p—n)(p—2n) -23+ &c. 2 2.3 =1+2x+P(p—n) +23+ &c. (18) n 2.n? 2.3.no If in this formula we make p=1, it becomes 1 1 ? (1+x)ñ=1==x+ (1—n)(1-2n) -23+ &c. 2.n? 2.3.ni (19) n Formula (18) may be expressed also in the following manner : (1+x)á P(p—n) z2___P(p—n)(p—2n) 23 + n n. 2n n. 2n. 3n P(pn)(p—2n)(p—3n) -X4+ &c. n. 2n. 3n. 4n (21) and a similar notation may be employed in (19) and (20). For example, to find the square root of 1+x; that is, to expand (1+x)} into a series. Making n=2 in (20), we find (1+x)}=1+ -204 + &c. 2.3.4.24 1 3. 5 1 3 23 or reducing the denominators as in (21), (1+z)&=1+36-2.6..++ 1. (a+s4=(1+) =a:(1+1+36-) - &e.) 2. ( 02)=(19+(1-3) --1 --&.) 8. (a+b) * =_*(1+) =a*(1+2 + P(p—m) " + P(p—n)(p—2n) 1: +&c.) 2.3.n3 |