2.2.6 Example F 73 The equation (k +1)yk+1 – kYk = k (2.35) can be written as A(kyk) = k. (2.36) Therefore, kyk = A-1(k) = ½k(k – 1) + A, (2.37) where A is an arbitrary constant. Dividing by k gives =...

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Extracted text: 2.2.6 Example F 73 The equation (k +1)yk+1 – kYk = k (2.35) can be written as A(kyk) = k. (2.36) Therefore, kyk = A-1(k) = ½k(k – 1) + A, (2.37) where A is an arbitrary constant. Dividing by k gives = A/k+ ½(k – 1). (2.38) Equation (2.35) can also be written as k Yk k +1 k Yk+1 (2.39) k +1 Therefore, k Pk = qk (2.40) %3D k +1 50 Difference Equations and k-1 1 II P: = T (2.41) k i=1 k-1 i k-1 Σ (i + 1) i +1 i=1 Ii Pr r=1 (2.42) k-1 k(k – 1) -Σ 2 i=1 Substitution of equations (2.41) and (2.42) into equation (2.9) gives equation (2.38).