Answer to Question #304668 in Discrete Mathematics for Tege

Question

12. Solve the recurrence relation a_n=a_n-1+n²,where a₀=7 by substitution

Answer 1

Let us solve the recurrence relation $a_n=a_{n-1}+n^2,$ where $a_0=7.$

The characteristic equation $k-1=0$ has the root $k=1,$ and hence the general solutionis of the form:

$a_n=C\cdot 1^n+b_n,$ where $b_n=n(an^2+bn+c)=an^3+bn^2+cn.$

It follows that

$an^3+bn^2+cn=a(n-1)^3+b(n-1)^2+c(n-1)+n^2.$

For $n=0$ we get $0=-a+b-c,$ and hence $a=b-c.$

For $n=1$ we get $a+b+c=1,$ and thus $b-c+b+c=1,$ that is $b=\frac{1}2.$ Therefore, $a=\frac{1}2-c.$

For $n=-1$ we get $-a+b-c=-8a+4b-2c+1.$

It follows that

$c-\frac{1}2+\frac{1}2-c=8c-4+2-2c+1,$ and therefore $6c=1.$

We conclude that $c=\frac{1}6,\ a=\frac{1}2-\frac{1}6=\frac{1}3.$

Consequently, the general solutionis of the form:

$a_n=C+\frac{1}3n^3+\frac{1}2n^2+\frac{1}6n.$

It follows that $7=a_0=C,$ and we conclude that the particular solution is the following:

$a_n=7+\frac{1}3n^3+\frac{1}2n^2+\frac{1}6n.$