Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/undated/prelims/mathematics-iii-resit/5/a/i/solution

For the homogeneous linear recurrence relation, try . Its characteristic equation of a linear recurrence is
The distinct roots of a polynomial give the complementary terms . A constant cannot absorb the forcing because is already a root of a polynomial of the characteristic equation; try a particular solution . Substitution gives , hence . The general solution is .
The initial conditions give and . Hence , , and

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