Past exam of the mathematics course of the University of Oxford / Undated / Prelims / Mathematics III resit / 5 / a

For the homogeneous linear recurrence relation, try $u_n=r^n$. Its characteristic equation of a linear recurrence isThe distinct roots of a polynomial $1,-4$ give the complementary terms $A+B(-4{)}^n$. A constant cannot absorb the forcing because $1$ is already a root of a polynomial of the characteristic equation; try a particular solution $cn$. Substitution gives $c=-4c+1$, hence $c=1/5$. The general solution is $u_n=A+B(-4{)}^n+n/5$.

The initial conditions give $A+B=0$ and $A-4B=4/5$. Hence $A=4/25$, $B=-4/25$, and

The homogeneous characteristic equation of a linear recurrence isThe repeated characteristic root of a linear recurrence $-1$ supplies the independent terms $(-1{)}^n$ and $n(-1{)}^n$. A constant particular solution is $1/2$, since $1/2=-2(1/2)-1/2+2$. Thus $u_n=1/2+(A+Bn)(-1{)}^n$.

The scan prints $u_2=0$, not $u_1=0$. These initial conditions are consistent. They give $A=1/2$ from $u_0=1$, then $0=1/2+A+2B$, so $B=-1/2$. ThereforeIn particular $u_1=1/2$ and $u_2=0$. Direct substitution verifies the recurrence relation. This is an example of nonconsecutive initial values for a linear recurrence; replacing the printed subscript by 1 would change the answer.