Solve: an + an-1 function method. = 3n2″, ao = 0, using Generating

To solve the recurrence relation:

an + an-1 = 3n^2 , where a0 = 0, using generating functions, we can follow these steps:

1. Define the generating function A(x) for the sequence {an} as follows:

A(x) = a0 + a1x + a2x^2 + a3x^3 + …

2. Multiply the recurrence relation by x^n and sum over all n≥1:

∑(an + an-1)x^n = ∑3n^2 x^n

3. Simplify the left-hand side using the shift property of the summation:

∑anx^n + ∑an-1x^n = ∑3n^2 x^n

∑anx^n + x ∑an-1x^(n-1) = ∑3n^2 x^n

4. Replace ∑anx^n by A(x) and use the identity ∑anx^n = xA(x) – a0:

A(x) + xA(x) – a0 = ∑3n^2 x^n / (1-x)

5. Replace the sum on the right-hand side by its generating function:

A(x) + xA(x) – a0 = 3x^2 / (1-x)^3

6. Solve for A(x):

A(x) = 3x^2 / [(1-x)^3 (1+x)]

7. Use partial fraction decomposition to write A(x) in a more manageable form:

A(x) = A1/(1-x) + A2/(1-x)^2 + A3/(1-x)^3 + A4/(1+x)

where A1 = 1, A2 = 2, A3 = 2, and A4 = -2.

8. Expand each term using the binomial series:

A(x) = 1 + 2x + 3x^2 + 4x^3 + …

9. Finally, read off the coefficients to get the solution to the original recurrence relation:

an = n(n+1).