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Recurrence equation for calculating the first n terms of a recursion
Here is a recurrence equation that I would like to solve.
Equation :
$T(1) = 1$
$T(k) = 3T(k-1) – (k-2)^2$
My Attempt at solution
I guessed $T(k) = a*k + b$ where $a$ and $b$ are such that I can get this as a first order linear ODE. However, I do not have a solution for this equation (i.e. I do not have the constants $a$ and $b$).
Can someone help me out? I would love to know the constants of the equation.
A:
The recurrence can be easily transformed to a first order differential equation:
$$\frac{\mathrm{d}T}{\mathrm{d}k} = 3T(k-1) – (k-2)^2$$
This can be easily solved using the method of integrating factors, and you’ll get:
$$T(k) = 3^k\int_1^k\int_0^x(3t-x^2)^\frac{ -1}{2}\mathrm{d}t\mathrm{d}x$$
Which, after a bit of simplification, is:
$$\int_1^k\int_0^x(3t-x^2)^\frac{ -1}{2}\mathrm{d}t\mathrm{d}x = \int_1^k\left(\int_0^x(3t-x^2)^\frac{
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