Giving the values in the parentheses: -2, -1, 0, 1, 2, we have the following results inclosed X, X, X-4 X-3 +744-49(58 +159 +260 +119-174)-313+224, forming a series of the fourth order. The series of the third order is of the second, of the first, -793 equal differences, 9(+217101-141-293)-139+537; +784 +226(-116 -242 -152)+154+676; By substituting other values, as -3, -4, -5, —6, and +3, +4, +5, +6, &c., we may extend the top series to any length. To save the time and trouble of substituting consecutive numbers and calcu lating the result, the method of difference series is employed, thus: Substitute a number of consecutive values one more than the degree of the equation; the smallest numbers, being more easily substituted, are preferred. In the present example, substituting -2, -1, 0, 1, 2, we obtain that portion of the first series which is of the 3° order, included in brackets; from this, by subtracting its consecutive terms, the corresponding portions of the series of the 2° order, and so on; and, finally, the difference, 216. Using this difference, we may extend the top series at pleasure, according to the method in Art. 330. The roots of the equation lie between those numbers the substitutions of which produce unlike signs in the result; thus, in the above there is one root between 3 and 4, one between 1 and -2, one between 1 and 2, and one between 3 and 4. 334. There exists between the coefficients of two consecutive powers of x+a relations from which many useful consequences may be deduced. Suppose the mth power of x+a to be -2 xm +Aar+Ba2x2+Ca3r-3+, &c. Multiplying the polynomial by x+a, there results m+1+Aax+Ba2rm-1+Ca3rm-2+ ... ... From which we conclude that, to obtain the coefficient of any term of the (m+1)th power of x+a, it is only necessary to add to the coefficient of the term of the same rank in the mth power that of the preceding term. 335. According to this rule, we can form the coefficients of the successive powers of x+a, as may be seen in the following table : The first vertical column of this table is formed of the single number 1. The second column is formed of the number 1 written twice. We form the third column by placing at the side of each term in the second column the number obtained by adding it to the term above it; we find thus, for the first term of the third column 1+0 or 1; the second term is 1+1 or 2, and the third 0+1 or 1. The fourth column is deduced from the third in the same manner that that is from the second, and so on. The two terms of the second column may be considered as the coefficients of the first power of x+a. It results from the above rule that the terms of the third column are the coefficients of the development of (x+a)2, those of the fourth column of (x+a)3, &c. This table, which may be indefinitely extended, is called the Arithmetical Triangle of Pascal. 336. It is easy to see from the composition of the arithmetical triangle that the pth term of any horizontal line is the sum of the p first terms of the preceding horizontal line. Because if we consider, for example, the term 56, which is the sixth of the fourth line, this term is formed by adding the two numbers 21 and 35, which are placed at its left in the third and fourth lines; but the second of these two numbers, 35, is the sum of 15 and 20; the last number, 20, is the sum of 10 and 10, and the last number, 10, the sum of 6 and 4; finally, 4 is the sum of the two numbers 3 and 1; we have, therefore, 56=21+15+10+6+3+1. THE DIFFERENTIAL METHOD OF SUMMING SERIES. 337. Let a, b, c, d, e,... . be a series of terms, in which each term is less than the succeeding one; and, taking the successive differences, we have Putting d1, d2, dз, d4, ..... for the first terms of the first, second, third, fourth,.... differences, we have b-a d-3c+3b-a &c. =d1..b=a+ dı =d.c=a+2d1+ da =d3d=a+3d1+3d2+ dз &c. e-4d6c-4b+a=d4 ··· e=a+4d,+6d2+4d3+ds, Hence the (n+1)th term of the proposed series is evidently (n-1 a+ndi+n- -d2+ 1.2.3 and, therefore, the nth term is (by writing n-1 for n) be two series, of which the (n+1)th term of the latter is obviously the sum of n terms of the former; but the first terms of the first, second, third, fourth, ..... differences in the latter, are a, b-a-d1, c-2b+a=ds, d-3c+8b-a, da, &c.; hence the (n+1)th term of the latter series, or the sum of n terms of the former, is, by (1) in the last article, 0+na+ d1+ n(n-1) n(n−1)(n—2) n(n−1)(n—2)(n—3) 1.2 1.2.3 1.2.3.4 (1) To what is 1.2+2.3+3.4+4.5+... n(n+1) equal? 2, 6, 12, 20, 30, is the given series ; 4, 6, 8, 10, differences of the consecutive terms; 2, 2, 0, 0. Hence, a=2, d1=4, d2=2, and dз, d1, &c. =0; therefore n(n-1) d1+ S=na+ 2 n(n-1) (n-2) da; 2.3 =2n+2n(n-1)+}n(n−1)(n−2) =}n(n+1)(n+2). Proceed always in this way till the differences become the same.t (2) Find the sum of n terms of the series 1, 23, 33, 43, 53, &c. (6) Find the sum of 15 terms of the series 1, 4, 8, 13, 19, &c. POWERS OF THE TERMS OF PROGRESSIONS. 339. If all the terms of a geometrical progression a:aq:aq: aq3....aqn-1 are raised to the same power m, the result is the series a", amqm, amqm, amq3m.....amqTM(n−1), which is a geometrical progression, of which the first term is aTM, the ratio qTM, and the number of terms n. 340. If the terms of a progression by differences, whose first term is a and common difference d, be each raised to the mth power, we have *This is the dy of the former series, but the d2 of the latter. The terms of the formula (2), containing those orders of differences which become zero, like dз, d4, &c., in example 1, will all vanish, and the expression for S will be composed only of the preceding terms. These differences being not the same, the same powers of the terms of an arithmetical progression do not form an arithmetical progression. 341. To find the sum of the mth powers of an arithmetical progression. Let a.b.c.d...k.l be any arithmetical progression, of which the common difference is d. Then b=a+s, c=b+d, ...........l=k+d. Raising these equalities to the power m+1, (m+1)m am-182+, &c. bm+1=am+1+(m+1)amd+- 1.2 Adding all these equalities, suppressing the common terms in the two equal sums, viz., bm+1, cm+1, &c., and transposing am+1, we have [m+1 —am+1=(m+1)8(am+bTM....+kTM), The law of the unwritten terms is sufficiently apparent, and the series must evidently end with the term preceding that which contains the factor m—m or 0. By formula (1) the sum Sm can be found, when the sums of the inferior powers are known; for this purpose, make m=0, the formula gives S.; making m=1, it gives S1, and so on to the sum of the powers required. If the progression÷a.a+d.a+25.......... is replaced by 1.2.3....N (or the series of natural numbers from 1 to N), i. e., a=1, d=1, l=N, then formula (1) becomes Formula (3) expresses the sum of 1o+2°+3°.... to N terms, or of 1+1 +1... to N. EXAMPLES. (1) If m=0 and N=10, So=N=10. Formula (4) expresses the sum of 1+2+3....+N. (2) If m=1 and N=10, S1= 10(10+1) 110 2 = =55. Formula (5) expresses the sum of 12+22+32.....+No. 10 × 11 × 21 (3) If m=2 and N=10, S,= =385. 6 PILING OF BALLS AND SHELLS. 342. Balls and shells are usually piled in three different forms, called triangular, square, or rectangular, according as the figure on which the pile rests is triangular, square, or rectangular. (1) A triangular pile is formed by continued horizontal courses of balls or shells laid one above another, and these courses or rows are usually equilateral triangles whose sides decrease by unity from the bottom to the top row, which is composed simply of one shot. Denoting by N the number of balls contained in one side of the equilateral triangle which forms the base of the triangular pile, it is evident that the number of balls in the base will be expressed by 1+2+3... +N or S1, which by (4) is equal to N2+N |