Let $\{a_n\}_{n=0}^{\infty}$ be a sequence such that $a_0 = a_1 = 0$ and $a_{n+2} = 3a_{n+1} - 2a_n + 1, \forall n \ge 0$. Then $a_{25} a_{23} - 2a_{25} a_{22} - 2a_{23} a_{24} + 4a_{22} a_{24}$ is equal to: Options: (A) 483 (B) 528 (C) 575 (D) 624

### Analyzing the Setup The given recurrence relation is defined as: $$a_{n+2} = 3a_{n+1} - 2a_n + 1$$ To simplify this, we rearrange the terms by splitting $3a_{n+1}$ into $a_{n+1} + 2a_{n+1}$. Moving one $a_{n+1}$ to the left side yields: $$a_{n+2} - a_{n+1} = 2a_{n+1} - 2a_n + 1$$ Factoring the right side, we obtain: $$a_{n+2} - a_{n+1} = 2(a_{n+1} - a_n) + 1$$ By defining a new sequence $b_n = a_{n+1} - a_n$, we successfully transform the second-order recurrence into a first-order linear recurrence: $$b_{n+1} = 2b_n + 1$$ ### The Telescoping Dance To solve $b_{n+1} = 2b_n + 1$, we add $1$ to both sides to reveal a Geometric Progression: $$b_{n+1} + 1 = 2(b_n + 1)$$ Given the initial condition $b_0 = a_1 - a_0 = 0$, we find the general term for $b_n$: $$b_n + 1 = (b_0 + 1) \cdot 2^n = 2^n \implies b_n = 2^n - 1$$ To find $a_n$, we utilize the telescoping sum property $a_n = a_0 + \sum_{k=0}^{n-1} b_k$. Since $a_0 = 0$, we have: $$a_n = \sum_{k=0}^{n-1} (2^k - 1)$$ Splitting the summation, we get: $$a_n = \sum_{k=0}^{n-1} 2^k - \sum_{k=0}^{n-1} 1 = (2^n - 1) - n$$ Thus, the explicit formula for the sequence is: $$a_n = 2^n - n - 1$$ ### The Algebraic Masterstroke We are tasked with evaluating the expression: $$E = a_{25} a_{23} - 2a_{25} a_{22} - 2a_{23} a_{24} + 4a_{22} a_{24}$$ By grouping the terms, we can factorize the expression: $$E = a_{25}(a_{23} - 2a_{22}) - 2a_{24}(a_{23} - 2a_{22})$$ $$E = (a_{25} - 2a_{24})(a_{23} - 2a_{22})$$ Now, we evaluate the general form $a_n - 2a_{n-1}$ using our explicit formula: $$a_n - 2a_{n-1} = (2^n - n - 1) - 2(2^{n-1} - (n-1) - 1)$$ $$a_n - 2a_{n-1} = (2^n - n - 1) - (2^n - 2n + 2 - 2) = n - 1$$ ### Final Calculation Using the result $a_n - 2a_{n-1} = n - 1$, we calculate the two factors of $E$: For the first factor ($n=25$): $25 - 1 = 24$. For the second factor ($n=23$): $23 - 1 = 22$. The final value of the expression is: $$E = 24 \times 22 = 528$$ **The final answer is 528.**

JEE Main 2022 (29 July Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Sequence and Series: Let ${a_{n}}_{n = 0}^{\infty}$ be a sequence such that $a_{0} = a_{1} = 0$ and $a_{n + 2} = 3 a_{n + 1} - 2 a_{n} + 1, \forall n \geq 0$ . Then $a_{25} a_{23} - 2 a_{25} a_{22} - 2 a_{23} a_{24} + 4 a_{22} a_{24}$ is equal to:

Q: How did you know to group the terms in the target expression $E = a_{25} a_{23} - 2a_{25} a_{22} - 2a_{23} a_{24} + 4a_{22} a_{24}$?

Look at the coefficients! The expression has a structure $XY - 2XZ - 2WY + 4WZ$. This is a classic factorization pattern: $X(Y - 2Z) - 2W(Y - 2Z) = (X - 2W)(Y - 2Z)$. Whenever you see such a symmetric structure in a sequence problem, always look for common factors first.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Don't rush to calculate the terms

a_{25}, a_{23}, a_{22}, a_{24}

individually. The recurrence relation

a_{n + 2} = 3 a_{n + 1} - 2 a_{n} + 1

is a classic linear recurrence. Try to rewrite it by grouping terms to see if you can isolate the difference between consecutive terms,

a_{n + 1} - a_{n}

Hint 2

Hint 2: The Arsenal. Define a new sequence

b_{n} = a_{n + 1} - a_{n}

. Substitute this into your rearranged recurrence relation. You will find that

b_{n + 1}

is related to

b_{n}

in a way that makes

b_{n}

a simple geometric progression after a small shift. This is your key to finding the explicit formula for

a_{n}

Hint 3

Hint 3: The Execution. Once you have the explicit formula

a_{n} = 2^{n} - n - 1

, look at the target expression

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

. Factor it by grouping. You will notice it simplifies to the product of two terms, each of the form

a_{k} - 2 a_{k - 1}

. Evaluate this general form

a_{n} - 2 a_{n - 1}

to find the final answer.

Visualized Solution

Given Sequence and Recurrence

Given sequence: ${a_{n}}_{n = 0}^{\infty}$
Initial conditions: $a_{0} = 0, a_{1} = 0$
Recurrence relation: $a_{n + 2} = 3 a_{n + 1} - 2 a_{n} + 1$

Rearranging the Recurrence

Rewrite the recurrence: $a_{n + 2} - a_{n + 1} = 2 a_{n + 1} - 2 a_{n} + 1$
Group terms: $a_{n + 2} - a_{n + 1} = 2 (a_{n + 1} - a_{n}) + 1$

Defining a Difference Sequence $b_{n}$

Let $b_{n} = a_{n + 1} - a_{n}$
The relation becomes: $b_{n + 1} = 2 b_{n} + 1$
Calculate initial term: $b_{0} = a_{1} - a_{0} = 0 - 0 = 0$

Converting to a Geometric Progression

Add $1$ to both sides: $b_{n + 1} + 1 = 2 b_{n} + 2$
Factor out $2$ : $b_{n + 1} + 1 = 2 (b_{n} + 1)$
This is a G.P. for the sequence $(b_{n} + 1)$ with common ratio $2$

General Term of $b_{n}$

General term of G.P.: $b_{n} + 1 = (b_{0} + 1) \cdot 2^{n}$
Substitute $b_{0} = 0$ : $b_{n} + 1 = (0 + 1) \cdot 2^{n} = 2^{n}$
Result: $b_{n} = 2^{n} - 1$

Expressing $a_{n}$ using Telescoping Sum

Recall $b_{k} = a_{k + 1} - a_{k}$
Summing from $k = 0$ to $n - 1$ : $\sum_{k = 0}^{n - 1} b_{k} = a_{n} - a_{0}$
Since $a_{0} = 0$ , we get: $a_{n} = \sum_{k = 0}^{n - 1} (2^{k} - 1)$

Calculating the General Term $a_{n}$

Split the sum: $\sum_{k = 0}^{n - 1} 2^{k} - \sum_{k = 0}^{n - 1} 1$
Sum of G.P. part: $\frac{1 \cdot ( 2 ^{n} - 1 )}{2 - 1} = 2^{n} - 1$
Sum of constant part: $n \cdot 1 = n$
General term: $a_{n} = 2^{n} - n - 1$

Analyzing the Target Expression

Target Expression: $E = a_{25} a_{23} - 2 a_{25} a_{22} - 2 a_{23} a_{24} + 4 a_{22} a_{24}$
Group the first two terms and the last two terms.
$E = a_{25} (a_{23} - 2 a_{22}) - 2 a_{24} (a_{23} - 2 a_{22})$

Factorizing the Expression

Common factor: $(a_{23} - 2 a_{22})$
Factorized form: $E = (a_{25} - 2 a_{24}) (a_{23} - 2 a_{22})$
Notice the recurring pattern: $(a_{n} - 2 a_{n - 1})$

Evaluating $(a_{n} - 2 a_{n - 1})$

Substitute $a_{n} = 2^{n} - n - 1$
Find $a_{n - 1}$ : $a_{n - 1} = 2^{n - 1} - (n - 1) - 1 = 2^{n - 1} - n$
Multiply by 2: $2 a_{n - 1} = 2 (2^{n - 1} - n) = 2^{n} - 2 n$

Simplifying the General Factor

Substitute into expression: $a_{n} - 2 a_{n - 1} = (2^{n} - n - 1) - (2^{n} - 2 n)$
Distribute negative sign: $2^{n} - n - 1 - 2^{n} + 2 n$
Cancel $2^{n}$ terms: $- n - 1 + 2 n$
Final simplified form: $a_{n} - 2 a_{n - 1} = n - 1$

Final Substitution and Calculation

For $n = 25$ : $a_{25} - 2 a_{24} = 25 - 1 = 24$
For $n = 23$ : $a_{23} - 2 a_{22} = 23 - 1 = 22$
Final value: $E = 24 \times 22$
Calculation: $24 \times 22 = 528$

00:00 / 00:00

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The given recurrence relation is defined as:

a_{n + 2} = 3 a_{n + 1} - 2 a_{n} + 1

To simplify this, we rearrange the terms by splitting

3 a_{n + 1}

into

a_{n + 1} + 2 a_{n + 1}

. Moving one

a_{n + 1}

to the left side yields:

a_{n + 2} - a_{n + 1} = 2 a_{n + 1} - 2 a_{n} + 1

Factoring the right side, we obtain:

a_{n + 2} - a_{n + 1} = 2 (a_{n + 1} - a_{n}) + 1

By defining a new sequence

b_{n} = a_{n + 1} - a_{n}

, we successfully transform the second-order recurrence into a first-order linear recurrence:

b_{n + 1} = 2 b_{n} + 1

The Telescoping Dance

To solve

b_{n + 1} = 2 b_{n} + 1

, we add

1

to both sides to reveal a Geometric Progression:

b_{n + 1} + 1 = 2 (b_{n} + 1)

Given the initial condition

b_{0} = a_{1} - a_{0} = 0

, we find the general term for

b_{n}

b_{n} + 1 = (b_{0} + 1) \cdot 2^{n} = 2^{n} ⟹ b_{n} = 2^{n} - 1

To find

a_{n}

, we utilize the telescoping sum property

a_{n} = a_{0} + \sum_{k = 0}^{n - 1} b_{k}

. Since

a_{0} = 0

, we have:

a_{n} = k = 0 \sum n - 1 (2^{k} - 1)

Splitting the summation, we get:

a_{n} = k = 0 \sum n - 1 2^{k} - k = 0 \sum n - 1 1 = (2^{n} - 1) - n

Thus, the explicit formula for the sequence is:

a_{n} = 2^{n} - n - 1

The Algebraic Masterstroke

We are tasked with evaluating the expression:

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

By grouping the terms, we can factorize the expression:

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

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

Now, we evaluate the general form

a_{n} - 2 a_{n - 1}

using our explicit formula:

a_{n} - 2 a_{n - 1} = (2^{n} - n - 1) - 2 (2^{n - 1} - (n - 1) - 1)

a_{n} - 2 a_{n - 1} = (2^{n} - n - 1) - (2^{n} - 2 n + 2 - 2) = n - 1

Final Calculation

Using the result

a_{n} - 2 a_{n - 1} = n - 1

, we calculate the two factors of

E

: For the first factor (

n = 25

25 - 1 = 24

. For the second factor (

n = 23

23 - 1 = 22

The final value of the expression is:

E = 24 \times 22 = 528

The final answer is 528.

Common Traps & FAQs

Wait, why can't I just calculate

a_{0}, a_{1}, a_{2}, \dots

until

a_{25}

? It seems straightforward.

While calculating terms one by one is possible, it is highly inefficient and prone to arithmetic errors in a JEE Advanced setting. The problem is designed to test your ability to recognize patterns and use algebraic manipulation to simplify complex expressions. Finding the general term

a_{n}

is the elegant, professional way to solve it.

How did you know to group the terms in the target expression

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

Look at the coefficients! The expression has a structure

X Y - 2 X Z - 2 W Y + 4 W Z

. This is a classic factorization pattern:

X (Y - 2 Z) - 2 W (Y - 2 Z) = (X - 2 W) (Y - 2 Z)

. Whenever you see such a symmetric structure in a sequence problem, always look for common factors first.

Is there a general way to solve

a_{n + 2} = A a_{n + 1} + B a_{n} + C

Yes, there is! You can use the method of undetermined coefficients or characteristic equations. However, for this specific problem, the 'difference method' (defining

b_{n} = a_{n + 1} - a_{n}

) is much faster and more intuitive. It reduces the second-order recurrence to a first-order one, which is much easier to handle.

Animated Solution for Mathematics - Sequence and Series: Let ${a_{n}}_{n = 0}^{\infty}$ be a sequence such that $a_{0} = a_{1} = 0$ and $a_{n + 2} = 3 a_{n + 1} - 2 a_{n} + 1, \forall n \geq 0$ . Then $a_{25} a_{23} - 2 a_{25} a_{22} - 2 a_{23} a_{24} + 4 a_{22} a_{24}$ is equal to:

Select Answer:

Visualized Solution

Given Sequence and Recurrence

Rearranging the Recurrence

Defining a Difference Sequence $b_{n}$

Converting to a Geometric Progression

General Term of $b_{n}$

Expressing $a_{n}$ using Telescoping Sum

Calculating the General Term $a_{n}$

Analyzing the Target Expression

Factorizing the Expression

Evaluating $(a_{n} - 2 a_{n - 1})$

Simplifying the General Factor

Final Substitution and Calculation

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Telescoping Dance

The Algebraic Masterstroke

Final Calculation

Similar Questions

Let ${a_{n}}_{n = 1}^{\infty}$ be a sequence such that $a_{1} = 1, a_{2} = 1$ and $a_{n + 2} = 2 a_{n + 1} + a_{n}$ for all $n \geq 1$ . Then the value of $47 \sum_{n = 1}^{\infty} \frac{a _{n}}{2 ^{3 n}}$ is equal to

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Let $\sum_{n = 0}^{\infty} \frac{n ^{3} (( 2 n )!) + ( 2 n - 1 ) ( n !)}{( n !) (( 2 n )!)} = a e + \frac{b}{e} + c$ , where $a, b, c \in Z$ and $e = \sum_{n = 0}^{\infty} \frac{1}{n !}$ . Then $a^{2} - b + c$ is equal to

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Select Answer:

Visualized Solution

Given Sequence and Recurrence

Rearranging the Recurrence

Defining a Difference Sequence .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } bn​

Converting to a Geometric Progression

General Term of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } bn​

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } an​ using Telescoping Sum

Calculating the General Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } an​

Analyzing the Target Expression

Factorizing the Expression

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (an​−2an−1​)

Simplifying the General Factor

Final Substitution and Calculation

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Telescoping Dance

The Algebraic Masterstroke

Final Calculation

Similar Questions

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General Term of $b_{n}$

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Calculating the General Term $a_{n}$

Evaluating $(a_{n} - 2 a_{n - 1})$

Let ${a_{n}}_{n = 1}^{\infty}$ be a sequence such that $a_{1} = 1, a_{2} = 1$ and $a_{n + 2} = 2 a_{n + 1} + a_{n}$ for all $n \geq 1$ . Then the value of $47 \sum_{n = 1}^{\infty} \frac{a _{n}}{2 ^{3 n}}$ is equal to

Let $⟨ a_{n} ⟩$ be a sequence such that $a_{0} = 0$ , $a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}$ , $n = 0, 1, 2, 3, ...$ Then $\sum_{k = 1}^{100} a_{k}$ is

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Let ${a_{n}}_{n = 0}^{\infty}$ be a sequence such that $a_{0} = a_{1} = 0$ and $a_{n + 2} = 2 a_{n + 1} - a_{n} + 1$ for all $n \geq 0$ . Then, $\sum_{n = 2}^{\infty} \frac{a _{n}}{7 ^{n}}$ is equal to

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