13. Find the last term of the harmonical series a, b, .. to n terms. 14. If m harmonical means are inserted between a and b, what is the second mean? 15. The first term of a harmonical series is x, and the second term is y; continue the series to three more terms. 16. The arithmetical mean between two numbers is and the harmonical mean is 42. What are the numbers ? 17. The fourth term of a harmonical series is the ninth term is. What is the seventh term? - 3, and 18. The geometrical mean between two numbers is 12, and the harmonical mean is 93. What are the numbers ? 33 40. 19. There are three numbers in harmonical progression whose sum is 23. If the second and third numbers are multiplied by 5 and 16, respectively, the three numbers form a geometrical progression. What are the numbers? 20. If the mth term of a harmonical progression is p, and the nth term is q, what is the (m+n)th term? 21. Prove that if a is the arithmetical mean between b and c, and b the geometrical mean between a and c, then e is the harmonical mean between a and b. XXX. THE BINOMIAL THEOREM. POSITIVE INTEGRAL EXPONENT. 442. The Binomial Theorem is a formula by means of which any power of a binomial may be expanded into a series. We shall consider in the present chapter those cases only in which the exponent is a positive integer. PROOF OF THE THEOREM FOR A POSITIVE INTEGRAL EXPONENT. 443. By actual multiplication, we obtain: (a + x)2= a2 + 2 ax + x2; (a + x)3 = a3 + 3'a2x +3 ax2 + x3; (a + x)= a1+4a3x+6a2x2+4ax + x; etc. In the above results we observe the following laws: I. The number of terms is greater by 1 than the exponent of the binomial. II. The exponent of á in the first term is the same as the exponent of the binomial, and decreases by 1 in each succeeding term. III. The exponent of x in the second term is 1, and increases by 1 in each succeeding term. IV. The coefficient of the first term is 1; and of the second term, is the exponent of the binomial. V. If the coefficient of any term is multiplied by the exponent of a in that term, and the result divided by the exponent of x increased by 1, the quotient will be the coefficient of the next following term. 444. We will now prove by induction (Art. 114, Note) that these laws hold for any positive integral power of a +2. Assume the laws to hold for (a + x)”, where n is any positive integer. Let P, Q, and R denote the coefficients of the terms involving a"-"x", a"--"+1, and a"-2a+2, respectively, in the second member of (1); thus, Multiplying both members by a +x, we have .... (2) ... +(P+Q)a"-"x"+1 + (Q+R) a"--1x"+2+ ··.· · (3) This result is in accordance with the second, third, and fourth laws of Art. 443. Since the fifth law of Art. 443 is assumed to hold with respect to the second member of (2), we shall have But nr is the exponent of a in that term of (3) whose coefficient is P+Q, and r+2 is the exponent of x increased by 1. Therefore the fifth law holds with respect to (3). Hence, if the laws of Art. 443 hold for any power of ax whose exponent is a positive integer, they also hold for a power whose exponent is greater by 1. But the laws have been shown to hold for (a + x)', and hence they also hold for (a + x)5; and since they hold for (a + x), they also hold for (a + x); and so on. Therefore the laws hold when the exponent is any positive integer. By aid of the fifth law, the coefficients of the successive terms after the third, in the second member of (2), may be readily found; thus, Note. In place of the denominators 1.2, 1.2.3, etc., it is cus- tomary to write [2, 3, etc. fies the product of the natural numbers from 1 to n inclusive. 445. Putting a = 1 in equation (4), Art. 444, we obtain EXAMPLES. 446. In expanding expressions by the Binomial Theorem, it is convenient to obtain the exponents and coefficients of the terms by aid of the laws of Art. 443, which have been proved to hold for any positive integral exponent. 1. Expand (a + x)3. The exponent of a in the first term is 5, and decreases by 1 in each succeeding term. The exponent of x in the second term is 1, and increases by 1 in each succeeding term. The coefficient of the first term is 1; of the second term, 5; multiplying 5, the coefficient of the second term, by 4, the exponent of a in that term, and dividing the result by the exponent of x increased by 1, or 2, we have 10 as the coefficient of the third term; and so on. Hence, (a + x) = a3 +5a1x+10a3x2 + 10 a2x2+5 ax1 +x3. Note 1. The coefficients of terms equally distant from the beginning and end of the expansion are equal. Thus the coefficients of the latter half of an expansion may be written out from the first half. 2. Expand (1 + 2 x1)®. (1 + 2 x1) = [1 +(2x1)]® = 1° 6.15.(2x) +15.1'.(2x1)2+20.13· (2 x1) 3 +15.12.(2x)+ 6.1.(2x)+(2x) = 1 + 12 x1 + 60 x3 + 160 x12 + 240 x16 +1920 +64 224. Note 2. If the first term of the binomial is a number expressed in Arabic numerals, it is convenient to write the exponents at first without reduction. The result should afterwards be reduced to its simplest form. Note 3. If either term of the binomial has a coefficient or exponent other than unity, it should be enclosed in a parenthesis before apply ing the laws. 3. Expand (3m — Vn). (3m - √n)' = [(3 m ̄ 1 ) + ( − n } ) ]• = · (3 m ̄ 1) + + 4 (3 m ̄ 2) 3 ( — n1) +6(3 m ̄1)2 ( — n3)? +4(3m ̄1) ( − n1)3 + ( − n3) • 1 1 = 81 m 2 — 108 m ̄ {n} +54 m ̄1n3 — 12 m ̄1n+n3. |