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,forall n≥ 0. Then a_25 a_23-2a_25 a_22-2a_23 a_24+4a_22 a_24 is equal to

Given,

$${a}_{0}=0,{a}_{1}=0$$a0​=0,a1​=0

And ${a}_{n+2}=3{a}_{a+1}-2{a}_{n+1}:n\geq 0$an+2​=3aa+1​−2an+1​:n≥0

$$\Rightarrow {a}_{n+2}-{a}_{n+1}=2({a}_{n+1}-{a}_{n})+1$$⇒an+2​−an+1​=2(an+1​−an​)+1

Now for $n=0{a}_{2}-{a}_{1}=2({a}_{1}-{a}_{0})+1....(1)$n=0a2​−a1​=2(a1​−a0​)+1....(1)

For $n=1{a}_{3}-{a}_{2}=2({a}_{2}-{a}_{1})+1........(2)$n=1a3​−a2​=2(a2​−a1​)+1........(2)

For $n=2{a}_{4}-{a}_{3}=2({a}_{3}-{a}_{2})+1........(3)$n=2a4​−a3​=2(a3​−a2​)+1........(3)

$$. . .$$...

For $n=n{a}_{n+2}-{a}_{n+1}=2({a}_{n+1}-{a}_{n})+1......(n)$n=nan+2​−an+1​=2(an+1​−an​)+1......(n)

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

$$({a}_{n+2}-{a}_{1})-2({a}_{n+1}-{a}_{0})-(n+1)=0$$(an+2​−a1​)−2(an+1​−a0​)−(n+1)=0

$$\Rightarrow {a}_{n+2}=2{a}_{n+1}+(n+1)$$⇒an+2​=2an+1​+(n+1)

Now replacing $n\rightarrow n-2$n→n−2 we get,

$$\Rightarrow {a}_{n}-2{a}_{n-1}=n-1$$⇒an​−2an−1​=n−1

Now putting $n=25$n=25 we get,

$$\Rightarrow {a}_{25}-2{a}_{24}=25-1=24$$⇒a25​−2a24​=25−1=24

Now putting $n=23$n=23 we get,

$$\Rightarrow {a}_{23}-2{a}_{22}=23-1=22$$⇒a23​−2a22​=23−1=22

So, $({a}_{25}-2{a}_{24})({a}_{23}-2{a}_{22})=(24)(22)=528$(a25​−2a24​)(a23​−2a22​)=(24)(22)=528