1.8.2 Example B Calculate the sum , k4. From number 5 of Table 1.1, for a = 1, b = 0, we obtain k(n+1) A-¼(n) (1.259) n +1 Now, from equation (1.220), for yk = k*, we have n A-1(n + 1)4 = E 59 +...

Explain the determaine

Extracted text: 1.8.2 Example B Calculate the sum , k4. From number 5 of Table 1.1, for a = 1, b = 0, we obtain k(n+1) A-¼(n) (1.259) n +1 Now, from equation (1.220), for yk = k*, we have n A-1(n + 1)4 = E 59 + constant, (1.260) r=1 or A-1(n + 1)4 + A, (1.261) r=1 where A is a constant whose value will be determined later. The term (n+1)4 is a polynomial of degree four and can be expressed as a sum of factorial functions: (n + 1)4 = co + cın(1) + c2n(2) + czn(3) + cạn(4) — Со + C1n + С2n(п — 1) + Сзп(п — 1)(п — 2) (1.262) + can (n - 1)(п - 2)(п — 3). THE DIFFERENCE CALCULUS 35 Setting n = 0 gives co = 1; setting n = 1 gives c1 + co = 16 and c1 = 15; setting n = 2 gives c2 = 25; setting n = 3 gives c3 = 10; finally, we obtain C4 = 1. Therefore, (n + 1)ª = n(4) + 10n(3) + 25n(2) + 15n(1) +1, (1.263) and applying A-1 to both sides and using equation (1.259) gives A-(n + 1)* = /sn(5) + 5/2n(4) + 25/3n (3) + 15/2n(2) + n), (1.264) or n Σ + 15/2n(2) + n(1) + A. /sn(5) + 5/2m(4) + 25/3n(3) (1.265) r=1 The constant A can be determined by taking n = 1 and obtaining 1 =1+A or A = 0. Finally, we have the result Ent = 1/sn(5) + 5/2n(4) + 25/3n(3) + 15/2n(2) + n(1). (1.266a) r=1 We can use equation (1.166) to express the factorial functions as powers of n; doing this gives n п(6п4 + 15п3 + 10n? — 1) (1.266b) 30 r=1