MISCELLANEOUS EXAMPLES IN FRACTIONS. 68. Though the treatment of algebraic fractions is only the adaptation of the arithmetical rules to general symbols, still it is necessary that the student should work a variety of examples in order to obtain that facility in their management which is so essential to his future progress in the higher branches of analysis. For exercise in all the operations on fractions, we shall give a few additional examples. Simplify as much as possible the subjoined examples : 69. INVOLUTION is the process of finding the powers of any quantity, and the operation is nothing else than multiplication, where the factors are all the same as the quantity itself. Since the nth power of x", or (x")" = x" × × × xTM to n factors. =xm+m+m+.... to n terms. and the mth power of x", or (x")" = x2 × x" × x” therefore the nth power of x is the same as the m'h (x)" = (x")TM = xmn. Hence to m factors. power of ", or 2018, (x2) = (x) = x*; (x3) = (x1)3 = x12, and also {(x + a)2 } = {(x + a)' }2 = (x + a)R. Thus any power of a power of a quantity is found by multiplying the indices of the two powers, for the index of the power required. = Also (2 x y3) = 2 x2 y3 × 2 x2 y3 = 4 x1 y° ; (3x) 2 = 3x 3 x 2y 3 ax = 9 a x2; and - 8x. The successive powers of a binomial quantity, a + b, may be found as follow: a3 + 2 a2b+ abs +ab+2ab+b3 a3 +3a2 b + 3 a b2 + b3 = (a + b)3 a* + 3 a3b+3a2b2 + a b3 a+4a3b+6a2 b2 + 4 a ba + bʻ = (a + b)‘. If this last product be multiplied by a + b, we shall have and multiplying this result by a + b, we get (a + b) = a® + 6 ab + 15 a1 b2 + 20 a3 b3 + 15 a2 b* + 6 a b3 + bo. In this manner any power of the binomial quantity a+b can be found; but by considering attentively the several terms of any of the preceding powers of a + b, we may deduce the law of the formation of the successive terms, and thence obtain the result of any power without the operation of multiplication. Thus the index of the leading quantity, a, in the first term, is always the index of the given power, and it decreases by unity from term to term to the last, where it is 0, and a being equivalent to 1 (40) is understood; hence -2 -3 are the n+1 factors involving the leading quantity, a. In like manner, since a and b are symmetrically involved in the quantity a + b, it is obvious that the powers of b will be the same as those of a, but in reverse order; hence bo, b', bo, b3, .... -3 bn 2, b" - 1, b", ...... (B) are the n+1 factors containing the second term, b, of the binomial a+b. Hence, recollecting that ao1 and 6o 1, the terms of (a+b)", without their coefficients, are the products of the corresponding terms of (a) and (8), viz.: -1 a", a" - 1 b, a" - * b3, a′′ - 3 b3, . . . . a3 b′′ − 3, a2 b′′ - 2, ab" - ', b" . . . . (y). With reference to the coefficients of the several terms, we observe that the coefficients of the first and last terms are each unity, the coefficients of the second term and the last but one are each the same as the index of the given power, and the coefficient of any term is found by multiplying the coefficient of the preceding term by the index of the power of the leading quantity in it, and dividing the product either by the number of terms to that place, or by the index of the power of the second quantity in it increased by unity; thus in the expansion of (a+b)' given above, we have 6 x 1 6 = 1; hence, generally, the coefficients of (a + b)" are n (n − 1) n (n − 1) n − 2 n (n − 1) (n − 2) n − 3 1, n, consequently the complete expression for (a + b)" is (a + b)" = a" + na"-1 b + n (n − 1) 2 n (n-1) (n-2) + 2.3 .... +b".... (1). etc.; where the last term is + b2 when ʼn is even, and b" when n is odd. This is the binomial theorem of NEWTON, which, as is shown in Art. 136, holds true whether n be a whole number or a fraction, and either positive or negative. The general term of the expansion of (a + b)" may be found by observing the law of formation of the several terms; thus the coefficient The whole number of terms is obviously one more than the index of the given power, and when the signs of both terms of the binomial are +, the signs of all the terms are +; but if the sign of the second term of the binomial is, the signs of the odd terms are +, and the signs of the even terms, or the signs of the terms are + and alternately. Also the coefficients equidistant from the extremes are equal, and the sum of the indices in any term is equal to the index of the power. EXAMPLES. 1. Find the square of a + b, and also the square of a b. α- b a + b a + b a2 + + ab + b2 (a+b)2= a+2ab+b2 These two forms are of frequent occurrence, and ought to be recollected by the student, as by means of them we can at once write down the squares of other binomial quantities; thus, if it be required to find the square of 2a + 3x, we have (2a+3x)2= (2 a)2 + 2 (2 a) (3x) + (3x)2 = 4a2 + 12 ax + 9x2; and (3x-7y)2 = (3 x)2 — 2 (3 ×x) (7 y) + (7 y)2 = 9x2-42xy +49 y2. a From these results we may also infer that in any trinomial which is a complete square, four times the product of the extreme terms is equal to the square of the middle term; since 4 x a2 × b2 = 4 a2 b2 = ( ± 2 ab)2. =2x, b=5y, Comparing this with the binomial (a + b)", we have a = and n=6; and as the fifth term is required, we must take p = 5, and by formula (3) we get the fifth term = 6 x 5 x 4 x 3 1.2.3.4 · (2x)2 (5 y) = 15 × 2 × 5 × x2y 37500 x2 y*. = × × 5* = ° 4. Find the square of 3 a3, and the cube or third power of 4 aR b3. Ans. 9 a' and 64 a® b9. 32 as x10 ys and a1o x3 y13. and the mth power of a x2 y3 z". 125 y° z3 16 a2 + 32 a x + 16x** 8. Find the fourth power of mx + ny. Ans. mx + 4 m3 n x3 ̧ 9. Find the sixth term of (a y+c z2) and the fourth term of (a-2x)*. 10. Find the fourth power of a Ans. 56 a3 c3 y3 z1o, and 2x and the cube of 1 Ans. a 8 a3 x + 24 a2x2 - 32 ax3 + 16 x*, and 1 6x + 21 x2 -32 ax3. 2x + 3x2. 44x3 +63 x* · 54 x5 +27 x6. x2, and the cube of 6 a2 32 a3 x + 24 a2 x6 8 a x + x2, - 540 a3 x + 450 a* x2 125 a3 3 12. Find the squares of a+b+c, and a2 a + t. Ans. a+b+c2 + 2 (a b + ac+bc), 2 a3 + § a2 — a + to. |