Скачать или смотреть a₀ = a₁ = 0 and aₙ₊₂ = 3 aₙ₊₁ − 2 aₙ + 1 for n ≥ 0; find a₂₅ a₂₃ − 2 a₂₅ a₂₂ − 2 a₂₃ a₂₄ + 4 a₂₂ a₂₄

📘 Math Problem Explanation
In this video we solve a recurrence-relation problem and evaluate a bilinear expression built from sequence terms — a compact algebra and sequences exercise useful for competitive exams.

👉 Complete Question:
Let the sequence
{
𝑎
𝑛
}
𝑛
=
0
∞
{a
n
​

}
n=0
∞
​

be such that
𝑎
0
=
𝑎
1
=
0
a
0
​

=a
1
​

=0 and, for all
𝑛
≥
0
n≥0,

𝑎
𝑛
+
2
=
3
𝑎
𝑛
+
1
−
2
𝑎
𝑛
+
1.
a
n+2
​

=3a
n+1
​

−2a
n
​

+1.

Then evaluate the expression

𝑎
25
 
𝑎
23
  
−
  
2
 
𝑎
25
 
𝑎
22
  
−
  
2
 
𝑎
23
 
𝑎
24
  
+
  
4
 
𝑎
22
 
𝑎
24
.
a
25
​

a
23
​

−2a
25
​

a
22
​

−2a
23
​

a
24
​

+4a
22
​

a
24
​

.

Options (example multiple choice):

483

528

575

624

🧩 Concepts covered:

Solving linear recurrence relations with constant coefficients (homogeneous plus particular solution)

Finding closed forms or useful relations between distant terms

Algebraic manipulation of products of sequence terms to simplify expressions

Exam strategies: using invariants, linearity, and telescoping-like identities

🎯 Who this helps:
Students preparing for IIT JEE, math Olympiads, or anyone practising recurrence relations and sequence identities.

Short SEO line: Solve a recurrence a_{n+2} = 3a_{n+1} − 2a_n + 1 and evaluate a bilinear expression of terms a₂₂ through a₂₅ — step-by-step.

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#Maths #IITJEE #RecurrenceRelations #Sequences #ProblemSolving #Algebra #CompetitiveExams