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Recurrence Relations

GATE CSE 2024 Set 2 · Q15

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Let T(n)T(n) be the recurrence relation defined as follows: T(0)=1,T(0) = 1, T(1)=2, andT(1) = 2, \text{ and} T(n)=5T(n−1)−6T(n−2)for n≥2T(n) = 5T(n - 1) - 6T(n - 2) \quad \text{for } n \ge 2 Which one of the following statements is TRUE?

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Solution

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Official answer: A) T(n) = (2n)

Recurrence Relations

To solve the linear homogeneous recurrence relation with constant coefficients:

T(n)−5T(n−1)+6T(n−2)=0T(n) - 5T(n-1) + 6T(n-2) = 0

The characteristic equation is:

r2−5r+6=0r^2 - 5r + 6 = 0
(r−2)(r−3)=0  ⟹  r=2,3(r - 2)(r - 3) = 0 \implies r = 2, 3

The general solution is:

T(n)=c1⋅2n+c2⋅3nT(n) = c_1 \cdot 2^n + c_2 \cdot 3^n

Using the given initial conditions:

  • For $n = 0$: $T(0) = c_1 + c_2 = 1$

  • For $n = 1$: $T(1) = 2c_1 + 3c_2 = 2$

Multiplying the first equation by 22 and subtracting from the second:

(2c1+3c2)−(2c1+2c2)=2−2  ⟹  c2=0(2c_1 + 3c_2) - (2c_1 + 2c_2) = 2 - 2 \implies c_2 = 0

Substituting c2=0c_2 = 0 into c1+c2=1c_1 + c_2 = 1 gives c1=1c_1 = 1.

Therefore, the closed-form solution is:

T(n)=1⋅2n+0⋅3n=2nT(n) = 1 \cdot 2^n + 0 \cdot 3^n = 2^n

Hence, T(n)=Θ(2n)T(n) = \Theta(2^n), which corresponds to (A).

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This Recurrence Relations previous year question is Q15 of the GATE CSE 2024 Set 2 paper. Find more GATE CSE PYQs on Recurrence Relations in Engineering Mathematics, or browse all GATE CSE PYQs.

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Recurrence Relations in GATE CSE

Recurrence Relations has 4 GATE CSE PYQs across 4 papers from 2016 to 2024. 3 of them are 1-mark MCQs like this one, and it is the only one in GATE CSE 2024 Set 2.

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GateAI solution cardGATE CSE 2024 Set 2 · Q15 · MCQ

Let T(n) be the recurrence relation defined as follows: T(0) = 1, T(1) = 2, and T(n) = 5T(n - 1) - 6T(n - 2) for n ≥ 2 Which one of the following statements is TRUE?

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