recurrence relation and series

Question

let $a_0$ and $a_1$ be arbitrary real number. for $n\geq2$ $$n(n-1)a_n=(n-1)(n-2)a_{n-1}-(n-3)a_{n-2}$$ find $a_0+a_1+...$

my book says that the answer is $a_1+a_0(e-1)$ where $e$ is $2.71828$ thing. I don't know how this appeared.

Answer 1

I fully agree with @saulspatz's comment. The problem is not simple and I then suppose that they almost ask you to use brute force.

Let $$a_n=\frac {(n-1)(n-2)a_{n-1}-(n-3)a_{n-2} }{n(n-1)}\qquad\text{with}\qquad a_0=\alpha\qquad a_1=\beta$$

This makes the first terms of the sequence to be $$\left\{\alpha ,\beta ,\frac{\alpha }{2},\frac{\alpha }{6},\frac{\alpha }{24},\frac{\alpha }{120},\frac{\alpha }{720},\frac{\alpha }{5040},\frac{\alpha }{40320},\frac{\alpha }{362880},\frac{\alpha }{3628800}\right\}$$ where you can recognize the factorials in denominators. Rewriting the sequence, it is $$\left\{\frac{\alpha }{1!} ,\beta ,\frac{\alpha }{2!},\frac{\alpha }{3!},\frac{\alpha }{4!},\frac{\alpha }{5!},\frac{\alpha }{6!},\frac{\alpha }{7!},\frac{\alpha }{8!},\frac{\alpha }{9!},\frac{\alpha }{10!}\right\}$$ Then $$\sum_{n=0}^p a_n=\beta+\alpha\sum_{n=0}^p \frac 1{n!}$$ Now, "remember" that the infinite expansion of the exponential function is $$e^x=\sum_{n=0}^\infty \frac {x^n}{n!}\implies e=\sum_{n=0}^\infty \frac {1}{n!}$$

Do you see the link ?