$a_{n}$ is a sequence of numbers defined by:
$a_{0}$ = −1
$a_{1}$ = 3
$a_{n+2}$ = 6$a_{n}$ − $a_{n+1}$ + 4n + 1
I have to prove that for every natural number n this equality is applicable:
$a_{n}$ = $2^{n}$ − $(-3)^{n}$ − n − 1
Any ideas/hints how to proceed? I have no idea what to do with that.
Sorry for bad English. That is not my first language.
Let's see three consecutive terms of the series and suppose that n is even:
$a_n=2^n - 3^n-n-1$
$a_{n+1}=2^{n+1} + 3^{n+1}-(n+1)-1$
$a_{n+2}=2^{n+2} -3^{n+2}-(n+2)-1$
Form the sum of $a_{n+1}$ and $a_{n+2}$:
$a_{n+1}+a_{n+2}=(2+4)2^n+(3-9)3^n-2n-5=6\overbrace{{(2^n-3^n-n-1)}}^{a_n}+4n+1$
$a_{n+2}=6a_n-a_{n+1}+4n+1$
We can do this for odd n, the result is same.