In the two examples which follow, the scale of relation consists of three parts. To find the value of S, from the equation S=A+B+C+ mxx (S-A-B)+nx2 x (S-A)+rx3 XS. Expanding, S =A+B+C+Smx-Amx - Bmx + Snx2- Anx2 + Srx3. Transp. & dividing, S= A+B+C-Amx-Bmx - Anx ̧ the formula. 1-mx-nx2 -rx3 Example 1. What is the sum of the infinite series, 1+4x +6x2+11x3+28x4 +63x5, &c.? = Let 1+4x+6x2 +11x3+28x4 +63x5 A+B+C+D+ E+F. By substituting numbers in the several formulæ, the scale of relation is found to be 2-1+3. Therefore, S= 1+4x+6x2-2x-8x2+x2 1+2x+x2-2x2 1-2x+x2-3x3 (1+x)2-2x2 (1 − x)2 — 3x 3° = = Example 2. What is the sum of the infinite series, 1+x +2x2+2x3-3x2 +3x5 +4x+4x7, &c.? Let 1+x+2x+2x3+3x2+3x5+4x+4x7 = A+B+C +D+E+F+G+H. By substituting in the several formulæ, the scale of relation is found to be 1+1-1. There 1+x+2x2-x — x2 — x2 fore S = 1-x-x2+x3 1 1-x-x2+x3 METHOD OF DIFFERENCES. Two examples under Article 493. f. Example 3. What is the 12th term of the series, 2, 6, 12, 20, 30, &c.? The given series= 2, 6, 12, 20, 30, The 1st order of differences=4, 6, 8, 10, = 2, 2, 2. The 2d order of differences= Making a, b, c, d, &c. the given series, and D', D", D"", &c. represent the first term, in the 1st, 2d, 3d, &c. orders of differences; a=2, D'=4, and_D"=2. Substituting num. bers in the formula, the 12th term=2+(12-1)4+(12-1)x Example 4. What is the 15th term of the series, 12, 22, 32, 42, 52, 62, &c.? The given series is, By expanding it becomes, 12, 22, 32, 42, 52, 62, 2, 2, 2, 2. The 1st order of differences = 3, 5, 7, 9, 11, The 2d order of differences= Making a, b, c, d, e, &c. the given series, and D', D", D"", &c.=the first term in the 1st, 2d, 3d, &c. orders of dif ferences, a=1, D'=3, and D"=2. Substituting numbers in the formula, the 15th term=1+(15−1;3+(15—1)× Five Examples under Article 493. g. Example 2. What is the sum of n terms of the series, 12, 22, 32, 42, 52, 62, 72, &c.? The given series is, 12, 22, 32, 4o, 5o, 6o, 72 By expanding it becomes, 1, 4, 9, 16, 25, 36, 49, 2, 2, 2, 2, 2. Making a, b, c, d, e, f, &c. the given series, and D', D", D"", the first term in the 1st, 2d, 3d, &c. orders of differences, a=1, D'=3, and D"=2. By the formula, the nth = {n(2n2 + 3n + 1) = fn(n+1)(2n+1.) Let n=20: then in(2n+3n+1)=120(800 +60 +1)=17220 = 2870. Example 3. What is the sum of n terms of the series, 13, 23, 33, 43, 53, &c.? The given series is 13, 23, 33, 43, 53, By expanding it becomes, 1, 8, 27, 64, 125, The 1st order of differences=7, 19, 37, 61, The 3d order of differences= 12, 18, 24, 6, 6. Making a, b, c, &c. and D', D", D"", &c. represent the series, and the 1st terms in the several orders of differences as in the previous examples, a=1, D'=7, D"=12, and D'" =6. By the formula, the nth term=n+n n X7+n 2 7n2-7n 24 12n3-36n2 +24n, 6n1-36n3 +66n2 —36n 24n, 84n2-84n, 48n3-144n2 +96n 2 24 24 + n1—6n3 +11n2 — 6n ____n1 +2n3+n2 n2 (n2+2n+1)=(nxn+1). Let n=50. +1))2=(25 × 51)2=1625625. == 4 (+50 × (50 Example 4. What is the sum of n terms of the series, 2, 6, 12, 20, 30, &c.? The given series is, 2, 6, 12, 20, 30, The 1st order of differences= 4, 6, 8, 10, The 2d order of differences= 2, 2, 2. Let a, b, c, &c. and D', D", &c. represent as in the preceding examples, a=2, D'=4, and D"-2. By the formu +2n)=3n(n2+3n+2)=3(n+1)×(n+2.) Example 5. What is the sum of 20 terms of the series, 1, 3, 6, 10, 15, &c.? The given series= 1, 3, 6, 10, 15 The 1st order of differences= 2, 3, 4, 5 The 2d order of differences= 1, 1, 1. Therefore a=1, D'=2, and D′′=1. By the formula, 20 Example 6. What is the sum of 12 terms of the series Therefore, a=1, D'=15, D"=50, D""=60, and D"""-24. Example 2. What are the roots of the equation x3 — 8x2 +4x+48-0? Let x=4.1 or 4.2. Then by substituting these numbers for x in the given equation, we have Here it will be observed, that the positive numbers are first added, and then subtracted from those which are negative. By the 1st supposition, the sum of the positive numbers is 133.321, which subtracted from -134.48, leaves -1.159. By the 2d supposition, the sum of the positive numbers is 138.888, which subtracted from -141.12, leaves -2.282. The difference between the errors -1.159, and -2.282, is-1.123. The value of .123 being small, it may be omitted in accordance with the usage of Day. (See the preceding example.) By the rule, (Art. 503.) −1.0 : 0.1-1.10.1, the correction. This subtracted from 4.1, leaves 4 for the value of x, one of the three roots. The others may be found by dividing according to Art. 462. x-4)x3 — 8x2+4x+48(x2 -4x-12 x2-4x=12. -12x+48 -12x+48 x2-4x+4=16. x2-2=±4. x=2+4=+ 6, or 4. Therefore the roots of the equation are -2, + 4, +6. Example 3. What are the roots of the equation a3+16x2 +65x-50-0? Let x=4.1 or 4.2. Then by substituting these numbers for x in the given equation, we have |