Let a_n_n=0^ be a sequence such that a_0=a_1=0 and a_n+2=3a_n+1-2a_n+1 for all n0.

Question

Let $\{a_n\}_{n=0}^{\infty}$ be a sequence such that $a_0=a_1=0$ and $a_{n+2}=3a_{n+1}-2a_n+1$ for all $n\ge0$. Then $a_{25}a_{23}-2a_{25}a_{22}-2a_{23}a_{24}+4a_{22}a_{24}$ is equal to:

Accepted Answer

The expression factors as $(a_{25}-2a_{24})(a_{23}-2a_{22})$.

From the recurrence, $a_{n+2}-2a_{n+1} = a_{n+1}-2a_n+1$. Letting $c_n=a_{n+1}-2a_n$, we get $c_{n+1}=c_n+1$ with $c_0=a_1-2a_0=0$, so $c_n=n$.

$$a_{25}-2a_{24}=c_{24}=24, \qquad a_{23}-2a_{22}=c_{22}=22.$$

$$24\times22 = 528.$$

Solutions · 1