Question

Let {an}ⁿ⁼⁰_∞ be a sequence such that a₀=a₁=0 and a_n+2=3a_n+1−2a_n+1,∀ n≥0. Then a₂₅a₂₃−2a₂₅a₂₂−2a₂₃a₂₄+4a₂₂a₂₄  is equal to

• A. 483
• B. 528
• C. 575
• D. 624

Answer 1

The Correct Option is B

a_n+2=3a_n+1−2a_n+1,∀ n≥0(a₀=a₁=0)
(a_n+2−a_n+1)−2(a_n+1−a_n)−1=0|
Put n = 0
(a₂−a₁)−2(a₁−a₀)−1=0
n = 1
(a₃−a₂)−2(a₂−a₁)−1=0
n = 2
(a₄−a₃)−2(a₃−a₂)−1=0
n=n
(a_n+2–a_n+1)−2(a_n+1−a_n)−1=0
On adding,
(a_n+2–a₁)−2(a_a+1−a₀)−(n+1)=0
∴a_n+2−2a_n+1−(n+1)=0
n\(\rightarrow\)n–2
a_n–2a_n−1−n+1=0
Now,
a₂₅a₂₃−2a₂₅a₂₂−2a₂₃a₂₄+4a₂₂a₂₄
=a₂₅(a₂₃−2a₂₂)−2a₂₄(a₂₃−2a₂₂)
=(a₂₅−2a₂₄)(a₂₃−2a₂₂)
=24⋅22
=528
So, the correct option is (B): 528

Answer 2

Consider the sequence \( \{a_n\}_{n=0}^{\infty} \) defined by the recurrence relation:

\( a_{n+2} - 3a_{n+1} + 2a_n = 1, \quad a_0 = a_1 = 0 \)

Step 1: Solve the Homogeneous Equation

The homogeneous part of the recurrence is:

\( a_{n+2} - 3a_{n+1} + 2a_n = 0 \)

Its characteristic equation is:

\( r^2 - 3r + 2 = 0 \)

Factoring, we get:

\( (r - 1)(r - 2) = 0 \)

Hence, the roots are \( r = 1 \) and \( r = 2 \), and the general solution of the homogeneous equation is:

\( a_n^{(h)} = A + B \cdot 2^n \)

Step 2: Find a Particular Solution

Since the non-homogeneous term is a constant (1), we try a particular solution of the form:

\( a_n^{(p)} = C \cdot n \)

Substituting into the recurrence, we have:

\( C(n+2) - 3C(n+1) + 2Cn = Cn + 2C - 3Cn - 3C + 2Cn = -C \)

Setting \( -C = 1 \) gives \( C = -1 \). Thus, the particular solution is:

\( a_n^{(p)} = -n \)

Step 3: General Solution and Initial Conditions

The general solution is the sum of the homogeneous and particular solutions:

\( a_n = A + B \cdot 2^n - n \)

Using the initial conditions:

• For \( n = 0 \): \( a_0 = A + B = 0 \)
• For \( n = 1 \): \( a_1 = A + 2B - 1 = 0 \)

Subtracting the first equation from the second yields:

\( (A+2B) - (A+B) = B = 1 \)

Therefore, \( A = -1 \), and the closed-form expression for the sequence is:

\( a_n = 2^n - n - 1 \)

Step 4: Evaluate the Expression

We wish to compute:

\( E = a_{25}a_{23} - 2a_{25}a_{22} - 2a_{23}a_{24} + 4a_{22}a_{24} \)

Notice that this expression factors as:

\( E = \bigl(a_{25} - 2a_{24}\bigr)\bigl(a_{23} - 2a_{22}\bigr) \)

Let’s simplify the term \( a_n - 2a_{n-1} \) using the closed form:

\( a_n - 2a_{n-1} = \Bigl(2^n - n - 1\Bigr) - 2\Bigl(2^{n-1} - (n-1) - 1\Bigr) \)

Expanding, we get:

\( = 2^n - n - 1 - 2^n + 2(n-1) + 2 = n - 1 \)

Therefore:

• For \( n = 25 \): \( a_{25} - 2a_{24} = 25 - 1 = 24 \)
• For \( n = 23 \): \( a_{23} - 2a_{22} = 23 - 1 = 22 \)

Thus, the expression simplifies to:

\( E = 24 \times 22 = 528 \)

Final Answer

The value of the expression is 528.