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JEE Main 2024
Sequences & Series
Sequences and Series
Hard

Question

 Let {an}n=0 be a sequence such that a0=a1=0 and an+2=3an+12an+1,n0.\begin{aligned} &\text { Let }\left\{a_{n}\right\}_{n=0}^{\infty} \text { be a sequence such that } a_{0}=a_{1}=0 \text { and } \\\\ &a_{n+2}=3 a_{n+1}-2 a_{n}+1, \forall n \geq 0 . \end{aligned} Then a25a232a25a222a23a24+4a22a24a_{25} a_{23}-2 a_{25} a_{22}-2 a_{23} a_{24}+4 a_{22} a_{24} is equal to

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Solution

Key Concepts and Formulas

  • Linear Non-Homogeneous Recurrence Relations: A recurrence relation of the form an+2+c1an+1+c0an=f(n)a_{n+2} + c_1 a_{n+1} + c_0 a_n = f(n). The general solution is the sum of the homogeneous solution and a particular solution.
  • Homogeneous Solution: For an+2+c1an+1+c0an=0a_{n+2} + c_1 a_{n+1} + c_0 a_n = 0, the characteristic equation is r2+c1r+c0=0r^2 + c_1 r + c_0 = 0. The form of the homogeneous solution depends on the roots of this equation.
  • Particular Solution: For a constant non-homogeneous term f(n)=kf(n) = k, if r=1r=1 is not a root of the characteristic equation, a particular solution is of the form an(p)=Ca_n^{(p)} = C. If r=1r=1 is a simple root, an(p)=Cna_n^{(p)} = Cn. If r=1r=1 is a double root, an(p)=Cn2a_n^{(p)} = Cn^2.
  • Algebraic Factoring: Simplifying complex expressions by grouping terms and finding common factors.

Step-by-Step Solution

Step 1: Analyze the Expression to be Evaluated The expression we need to evaluate is a25a232a25a222a23a24+4a22a24a_{25} a_{23} - 2 a_{25} a_{22} - 2 a_{23} a_{24} + 4 a_{22} a_{24}. We observe that this expression can be factored by grouping. a25a232a25a222a23a24+4a22a24=a25(a232a22)2a24(a232a22)a_{25} a_{23} - 2 a_{25} a_{22} - 2 a_{23} a_{24} + 4 a_{22} a_{24} = a_{25}(a_{23} - 2a_{22}) - 2a_{24}(a_{23} - 2a_{22}) =(a252a24)(a232a22)= (a_{25} - 2a_{24})(a_{23} - 2a_{22}) This factorization simplifies the problem significantly. Our goal is now to find the values of (an2an1)(a_{n} - 2a_{n-1}) for specific values of nn.

Step 2: Find the Closed-Form Expression for the Sequence ana_n The given recurrence relation is an+2=3an+12an+1a_{n+2} = 3a_{n+1} - 2a_n + 1. First, consider the homogeneous part: an+2=3an+12ana_{n+2} = 3a_{n+1} - 2a_n. The characteristic equation is r23r+2=0r^2 - 3r + 2 = 0. Factoring the quadratic, we get (r1)(r2)=0(r-1)(r-2) = 0. The roots are r1=1r_1 = 1 and r2=2r_2 = 2. The homogeneous solution is an(h)=A(1)n+B(2)n=A+B2na_n^{(h)} = A(1)^n + B(2)^n = A + B \cdot 2^n.

Next, find a particular solution for the non-homogeneous term '+1'. Since 1 is a root of the characteristic equation, we guess a particular solution of the form an(p)=Cna_n^{(p)} = Cn. Substitute this into the recurrence relation: C(n+2)=3C(n+1)2Cn+1C(n+2) = 3C(n+1) - 2Cn + 1 Cn+2C=3Cn+3C2Cn+1Cn + 2C = 3Cn + 3C - 2Cn + 1 Cn+2C=Cn+3C+1Cn + 2C = Cn + 3C + 1 2C=3C+12C = 3C + 1 C=1    C=1-C = 1 \implies C = -1. So, the particular solution is an(p)=na_n^{(p)} = -n.

The general solution is the sum of the homogeneous and particular solutions: an=an(h)+an(p)=A+B2nna_n = a_n^{(h)} + a_n^{(p)} = A + B \cdot 2^n - n.

Step 3: Use Initial Conditions to Determine Constants A and B We are given a0=0a_0 = 0 and a1=0a_1 = 0. For n=0n=0: a0=A+B200=A+B=0    A=Ba_0 = A + B \cdot 2^0 - 0 = A + B = 0 \implies A = -B. For n=1n=1: a1=A+B211=A+2B1=0a_1 = A + B \cdot 2^1 - 1 = A + 2B - 1 = 0. Substitute A=BA = -B into the second equation: B+2B1=0    B1=0    B=1-B + 2B - 1 = 0 \implies B - 1 = 0 \implies B = 1. Since A=BA = -B, we have A=1A = -1. Thus, the closed-form expression for the sequence is an=1+2nna_n = -1 + 2^n - n.

Step 4: Evaluate the Factored Expression We need to evaluate (a252a24)(a232a22)(a_{25} - 2a_{24})(a_{23} - 2a_{22}). Let's find a general form for (an2an1)(a_n - 2a_{n-1}). Using the closed form an=2nn1a_n = 2^n - n - 1: an2an1=(2nn1)2(2n1(n1)1)a_n - 2a_{n-1} = (2^n - n - 1) - 2(2^{n-1} - (n-1) - 1) =2nn12(2n1n+11)= 2^n - n - 1 - 2(2^{n-1} - n + 1 - 1) =2nn12(2n1n)= 2^n - n - 1 - 2(2^{n-1} - n) =2nn122n1+2n= 2^n - n - 1 - 2 \cdot 2^{n-1} + 2n =2nn12n+2n= 2^n - n - 1 - 2^n + 2n =n1= n - 1 So, an2an1=n1a_n - 2a_{n-1} = n - 1.

Now, we can substitute the values of nn into this result: For the first factor, (a252a24)(a_{25} - 2a_{24}), we have n=25n = 25. So, a252a24=251=24a_{25} - 2a_{24} = 25 - 1 = 24. For the second factor, (a232a22)(a_{23} - 2a_{22}), we have n=23n = 23. So, a232a22=231=22a_{23} - 2a_{22} = 23 - 1 = 22.

The expression is the product of these two values: (a252a24)(a232a22)=(24)(22)(a_{25} - 2a_{24})(a_{23} - 2a_{22}) = (24)(22) 24×22=24×(20+2)=24×20+24×2=480+48=52824 \times 22 = 24 \times (20 + 2) = 24 \times 20 + 24 \times 2 = 480 + 48 = 528

Common Mistakes & Tips

  • Incorrectly solving the characteristic equation or finding the particular solution: Double-check the roots and the form of the particular solution based on whether the non-homogeneous term's root is a root of the characteristic equation.
  • Algebraic errors in simplification: Carefully perform the substitutions and algebraic manipulations, especially when dealing with exponents.
  • Forgetting to use initial conditions: The constants A and B are crucial for obtaining the correct closed-form expression, which is necessary for the final calculation.
  • Not factoring the target expression: Trying to calculate individual ana_n terms and then substituting them into the original expression would be extremely tedious and error-prone.

Summary The problem involves a linear non-homogeneous recurrence relation. We first factored the expression to be evaluated into (a252a24)(a232a22)(a_{25} - 2a_{24})(a_{23} - 2a_{22}). Then, we found the closed-form expression for the sequence ana_n by solving the homogeneous and particular parts of the recurrence relation and using the initial conditions. We derived a simplified form for (an2an1)(a_n - 2a_{n-1}) as (n1)(n-1). Finally, we substituted the appropriate values of nn to calculate the two factors and their product, yielding the final answer.

The final answer is 528\boxed{528}.

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