Given,
a0=0,a1=0
And an+2=3aa+1−2an+1:n≥0
⇒an+2−an+1=2(an+1−an)+1
Now for n=0a2−a1=2\left({a}_{1} - {a}_{0}\right) + 1 . . . . (1)
For n=1a3−a2=2(a2−a1)+1........(2)</p><p>Forn = 2{a}_{4} - {a}_{3} = 2 \left({a}_{3} - {a}_{2}\right)+1........(3)
. . .
For n=nan+2−an+1=2(an+1−an)+1......(n)</p><p>Nowaddingallaboveequationuptonweget,</p><p>(an+2−a1)−2(an+1−a0)−(n+1)=0
⇒an+2=2an+1+(n+1)</p><p>Nowreplacingn \to n - 2weget,</p><p>⇒ {a}_{n} - 2{a}_{n - 1} = n - 1</p><p>Nowputtingn = 25weget,</p><p>⇒ {a}_{25} - 2{a}_{24} = 25 - 1 = 24</p><p>Nowputtingn = 23weget,</p><p>⇒ {a}_{23} - 2{a}_{22} = 23 - 1 = 22</p><p>So,(a25−2a24)(a23−2a22)=(24)(22)=528