Mathematics - Sequence & Series Question with Solution | TestHub

Given,

$a_0=0,a_1=0$ 

And $a_{n+2}=3a_{a+1}-2a_{n+1}:n\ge 0$

$\Rightarrow a_{n+2}-a_{n+1}=2\left(a_{n+1}-{\mathrm{a}}_n\right)+1$

Now for $n=0 a_2-a_1=2\left(a_1-a_0\right)+1 ....\left(1\right)$

For $n=1 a_3-a_2=2\left(a_2-a_1\right)+1 ........\left(2\right)$

For $n=2 a_4-a_3=2\left(a_3-a_2\right)+1 ........\left(3\right)$

$$\begin{gathered} . \\ . \\ . \end{gathered}$$

For $n=n a_{n+2}-a_{n+1}=2\left(a_{n+1}-a_n\right)+1 ......\left(n\right)$

Now adding all above equation upto $n$ we get,

$\left(a_{n+2}-a_1\right)-2\left(a_{n+1}-a_0\right)-\left(n+1\right)=0$

$\Rightarrow a_{n+2}=2a_{n+1}+\left(n+1\right)$

Now replacing $n\to n-2$ we get,

$\Rightarrow a_n-2a_{n-1}=n-1$

Now putting $n=25$ we get,

$\Rightarrow a_{25}-2a_{24}=25-1=24$

Now putting $n=23$ we get,

$\Rightarrow a_{23}-2a_{22}=23-1=22$

So, $\left(a_{25}-2a_{24}\right)\left(a_{23}-2a_{22}\right)=\left(24\right)\left(22\right)=528$