Find the square root of each of the following expressions : Find the cube root of each of the following expressions : 16. x-6-6x-5 + 12 x-1 — 8x-3. 17. 8 x — 36 x3 – 27 x2 + 54 x‡. х 18. 8x-3+12x-2 30 x 135+ 45 x + 27 x2 - 27 x3. 20. 8 x3y ̄2 + 13 x3 + y22 + 12 x3y−1 + 18 x2y ̄1⁄2 + 9 xy1 + 3 x1y. Solve each of the following equations: 21. 2x-2-x-1 1 = 0. 22. 2-% +52-1-11=0 24. 5x-2x2 - 16 – 0. 26. 5 x ̄ − 2 x ̄‡ – 3 − 0. = CHAPTER XXVII. PROGRESSIONS. § 1. 1. A Series is a succession of numbers, each formed according to some definite law. The single numbers are called the Terms of the series. The law may specify that each term shall be formed from the immediately preceding term in a prescribed way. E.g., in the series 1+3+5+7+9+... (1) each term after the first is formed by adding 2 to the preceding term. each term after the first is formed by multiplying the preceding term by 2. Or the law may state a definite relation between each term and the number of its place in the series. E.g., in the series 1+1+1 + 1 + ·· ... each term is the reciprocal of its number. (3) 2. The number of terms in a series may be either limited or unlimited. A Finite series is one of a limited number of terms. An Infinite series is one of an unlimited number of terms. In this chapter a few simple and yet very important series will be discussed. § 2. ARITHMETICAL PROGRESSION. 1. An Arithmetical Series, or as it is more commonly called an Arithmetical Progression (A. P.), is a series in which each term, after the first, is formed by adding a constant number to the preceding term. See § 1, Art. 1, (1). Evidently this definition is equivalent to the statement, that the difference between any two consecutive terms is constant. E.g., in the series we have 1+3+5+7+... 3-15-3 = 7 — 5 =..... For this reason the constant number of the first definition is I called the Common Difference of the series. 2. Let and a stand for the first term of the series, an for the nth (any) term of the series, S for the sum of n terms of the series. n The five numbers a, a,, d, n, S, are called the Elements of the progression. 3. The common difference may be either positive or negative. If d be positive, each term is greater than the preceding, and the series is called a rising, or an increasing progression. E.g., 1+ 2+ 3 + 4 + ···, wherein d A: If d be negative, each term is less than the preceding, and the series is called a falling, or a decreasing progression. E.g., 1-1-3 – 5 – ........, wherein d = - 2. 4. In an arithmetical progression any term is equal to the first term plus the product of the common difference and a number one less than the number of the required term, i.e., = a1 +2d, etc. By the definition of an arithmetical progression а1 = α1, а2 = a1+d, a2 = = a2+d: The law expressed by the formulæ for these first three terms is evidently general, and since the coefficient of d in each is one less than the number of the corresponding term, we have a1 = a1+ (n-1)d. Ex. 1. Find the 15th term of the progression we have therefore a2s=1+ (151)2=1+28 = 29. This formula may be used not only to find a,, when a1, d, and n are given, but also to find any one of the four numbers involved when the other three are given. Ex. 2. If a5= 3 (n = 5), and a1 = 1, we have 3=1+4d; whence d = 5. In any arithmetical progression, the sum of n terms is equal to one-half the product of the number of terms and the sum of the first and the nth term, i.e., Since the successive terms in an arithmetical progression, from the first to the nth inclusive, may be obtained either by repeated additions of the common difference beginning with the first term, or by repeated subtractions of the common difference beginning with the nth term, we may express the sum of n terms in two equivalent ways: S12=α1+(α2+d)+(α1+2d) + ··· +(α1+n−2·d)+(α1+n−1·d), ... ... Sn=an+(and)+(a,−2 d) + ··· + (a,n−2⋅d)+(a,n−1·d). Whence, by addition, ... 2 Sn = (a1+an) + (α1 + αn) + ··· + (α1 +an) + (α1 + αn), wherein there are n binomials a1 + a„. n Therefore, 2 S1 = n (α1 + ɑ„), or S1 = 1⁄2) (α1 + αn). 6. If the value of a given in (I.), be substituted for a, in (II.), we obtain Sn = · [2 a1 + (n − 1)d]. 2 (III.) Formula (II.) is used when a1, an, and n are given; and (III.) when a1, d, and n are given. 7. Ex. 1. If α1 = 1, α = 3, then S, = § (1 + 3) = 10. Ex. 2. If a1 =-4, d=2, n = 12, then 5 S12= 12 [2(−4) + 11 × 2] = 84. Either (II.) or (III.) can be used to determine any one of the five elements a1, an, d, n, Sn, when the three others involved in the formula are known. or Ex. 3. Given a1 -= n 3, d = 2, S2 = 12, to find n. From (III.), 12 = 2[−6 + 2 (n − 1)], n2 - 4 n = 12; whence n = 6 and - 2. The result 6 gives the series - 3−1+1+3+5+7, = 12. Since the number of terms must be positive, the negative result, -2, is not admissible. But its meaning may be assumed to be that two terms, beginning with the last and counting toward the first, are to be taken. 8. Formulæ (I.) and (II.), or (I.) and (III.), may be used simultaneously to determine any two of the five numbers a an, d, Sn, n, when the three others are given. Ex. 1. Given d = 1, n = 9, α, = 5, to find a1 and S. From (I.), 5 = a1 +8.1, whence a1: 1. From (II.), using the value of a just found, S=(1+5)=27. Ex. 2. Given a1 = 3, n = 13, S1 = 13, to find d and α13. 13 From (II.), 1313 (3+ a13), whence a1 =-1. From (I.), -1=3+12 d, whence d == Solving (1) and (2), we obtain n=12, a1 = 6; and n=5, a1 = -8. |