XLV. The rule for finding the coefficients of the powers of binomials may be derived and expressed more generally as follows: It is required to find the coefficients of the nth power of a + x. It has already been observed, Art. XLI., that the coefficient of the second term of the nth power is the nth term of the series of the second order, 1, 2, 3, &c., or, the sum of n terms of the series 1, 1, 1, &c.; that the coefficient of the third term is the sum of (n 1) terms of the series of the second order; that the coefficient of the fourth term is the sum of (n − 2) terms of the series of the third order, &c. So that the coefficient of each term is the sum of a number of terms of the series of the order less by one, than is expressed by the place of the term ; and the number of terms to be used is less by one for each succeeding series. By Art. XLII. the sum of n terms of the series 1, 1, 1, is i The sum of (n − 1). terms of the series of the second order is n (n 1 X 2 The sum of (n - 2) terms of the series of the third order is n (n · 1) (n 1) 2) 2) n (n 1) (n + a n-3 200 + &c. 1 X 2 X 3 It may be observed that n is the exponent of a in the first term, and that n or its equal , forms the coefficient of the second term. 1 nl or 1) is the The coefficient of the third term is multiplied by a multiplied by (n − 1) and divided by 2. But (n exponent of a in the 2d term, and 2 marks the place of the second term from the left. Therefore the coefficient of the third term is found by multiplying the coefficient of the second term by the exponent of a in that term, and dividing the product by the number which marks the place of that term from the left. By examining the other terms, the following general rule will be found true. Multiply the coefficient of any term by the exponent of the leading quantity in that term, and divide the product by the number that marks the place of that term from the left , and you will obtain the coefficient of the next succeeding term. Then diminish the exponent of the leading quantity by 1 and increase that of the other by 1 and the term is complete. By this rule only the requisite number of terms can be obtained. For 2", which is properly the last term of (a + x)", is the same as ao x. If we attempt by the rule to obtain another term from this, it becomes 0 x a-xc nt' which is equal to zero. It has been remarked above, that the coefficients of the last half of the terms of any power, are the same as those of the first reversed. This may be seen from the general expression: This furnishes the following fractions, viz. 1, i, j, 4, 3, 4, +. The first of these is the coefficient of the second term ; the coefficient of the second multiplied by i forms the coefficient of the third term, &c. ixf = 21. 21 X = 35. Now 35 multiplied by = 1 will not be altered.; hence two successive coefficients will be alike. 21 multiplied by produced 35; so 35 multiplied by must reproduce 21. In this way all the terms will be reproduced; for the last half of the fractions are the first half inverted. This demonstration might be made more general, but it is not necessary. XLVI. Progression by Difference, or Arithmetical Progression. A series of numbers increasing or decreasing by a constant difference, is called a progression by difference, and sometimes an arithmetical progression. The first of the two following series is an example of an increasing, and the second of a decreasing, progression by difference. 5, 8, 11, 14, 17, 20, 23....... 50, 45, 40, 35, 30, 25, 20....... It is easy to find any term in the series without calculating the intermediate terms, if we know the first term, the common difference, and the number of that term in the series reckoned from the first. Let a be the first term, r the common difference, and n the number of terms. The series is a, a tr, a + 2r, a + 3r....a + (n − 2)r, a + (n − 1)r. The points ... are used to show that some terms are left out of the expression, as it is impossible to express the whole until a particular value is given to n. Let l be the term required, then l=a + (n − 1)r. Hence, any term may be found by adding the product of the common difference by the number of terms less one, to the first term. Example. What is the 10th term of the series 3, 5, 7, 9, &c. 1=3 + 93 x 2 = 21. In a decreasing series, r is negative. Example. What is the 13th term of the series 48, 45, 42, &c. ? a = 48, r = -3, and n -1 = 12. I=48+ (12 X-3)= 48 - 36 = 12. b-a=cb a tc 2 That is, if three successive terms in a progression by difference be taken, the sum of the extremes is equal to twice the b= = mean. Example. Let the three terms be 3, 5, and 7. 2 X 5=7+3= 10. Example 2d. Let 7 and 17 be the first and last term, what is the mean? 7+17 = 12. Let a, b, c, d, be four successive terms of a progression by difference. ba=d-C b+c=a+d. That is, the sum of the two extremes is equal to the sum of the two means. Example. Let 5, 9, 13, 17, be four successive terms. 9 + 13 = 17 +5 = 22. Let a, b, c, d, e..... h, i, k, l, be any number of terms in a progression by differences; by the definition we have b-a=cb=d-ered=i-h=k-i=1-k. ba=1-K d-c=i-h, &c. which by transposition give a+1=b+k, c+1 = d + b, c. That is, if the first and last be added together, the second and the last but one, the third and the last but two, the sums will all be equal. Example. Let 3, 5, 7, 9, 11, 13, be such a series. |