Consider the following recurrence relation: T(n) = 2T(n - 1) + n2n for n > 0, T(0) = 1. Which ONE of the following options is CORRECT? A) T(n) = (n2 2n) B) T(n) = (n 2n) C) T(n) = ((log n)2 2n) D) T(n) = (4n)

Official answer: A) T(n) = (n2 2n). Explanation: To solve the recurrence relation: T(n) = 2T(n - 1) + n2n Divide both sides by 2n: (T(n))/(2n) = T(n - 1)2n-1 + n Let S(n) = (T(n))/(2n). Then the base condition becomes: S(0) = (T(0))/(20) = 1 Now, rewrite the recurrence in terms of S(n): S(n) = S(n - 1) + n Unrolling the recurrence: S(n) = S(0) + Σi=1n i = 1 + (n(n + 1))/(2) = (n2) Multiplying back by 2n gives: T(n) = 2n S(n) = 2n ( 1 + (n(n + 1))/(2) ) = (n2 2n) Hence, the correct option is (A).

Asymptotic Analysis

GATE CSE 2025 Set 1 · Q20

MCQ1 markEasy

Time00:00

Consider the following recurrence relation: $T(n) = 2T(n - 1) + n2^n \text{ for } n > 0, \quad T(0) = 1.$ Which ONE of the following options is CORRECT?

Select one answerSingle correct

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Solution

Step-by-step verified solution

Official answer: A) T(n) = (n2 2n)

Asymptotic Analysis

To solve the recurrence relation:

T(n) = 2T(n - 1) + n2^n

Divide both sides by $2^n$ :

\frac{T(n)}{2^n} = \frac{T(n - 1)}{2^{n-1}} + n

Let $S(n) = \frac{T(n)}{2^n}$ . Then the base condition becomes:

S(0) = \frac{T(0)}{2^0} = 1

Now, rewrite the recurrence in terms of $S(n)$ :

S(n) = S(n - 1) + n

Unrolling the recurrence:

S(n) = S(0) + \sum_{i=1}^{n} i = 1 + \frac{n(n + 1)}{2} = \Theta(n^2)

Multiplying back by $2^n$ gives:

T(n) = 2^n S(n) = 2^n \left( 1 + \frac{n(n + 1)}{2} \right) = \Theta(n^2 2^n)

Hence, the correct option is (A).

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This Asymptotic Analysis previous year question is Q20 of the GATE CSE 2025 Set 1 paper. Find more GATE CSE PYQs on Asymptotic Analysis in Algorithms, or browse all GATE CSE PYQs.

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Asymptotic Analysis in GATE CSE

Asymptotic Analysis has 31 GATE CSE PYQs across 19 papers from 2011 to 2026. 15 of them are 1-mark MCQs like this one, and it is the only one in GATE CSE 2025 Set 1.

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Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.