Problem 53 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(f(n+1)=\frac{2 f(n)+1}{2}, n=1,2, \ldots\) and \(f(1)=2\), then \(f(101)\) equals (A) 52 (B) 49 (C) 48 (D) 51

Step-by-Step Solution

Answer

(A) 52

Step 1: Understand the Recurrence Relation

The problem gives a recurrence relation: \( f(n+1) = \frac{2f(n) + 1}{2} \). This means to find \( f(n+1) \), you apply the formula using \( f(n) \). We start from \( f(1) = 2 \).

Step 2: Calculate Next Few Terms

To better understand the sequence, calculate the next few terms.\( f(2) = \frac{2f(1) + 1}{2} = \frac{2(2) + 1}{2} = \frac{5}{2} = 2.5 \). Similarly,\( f(3) = \frac{2(2.5) + 1}{2} = \frac{6}{2} = 3 \).

Step 3: Identify Pattern or Rule

Continuing this pattern: \( f(4) = \frac{2(3) + 1}{2} = 3.5 \), \( f(5) = 4 \). It seems that \( f(n) \) increments by 0.5 each step. Test further to confirm.

Step 4: Generalize the Pattern

Seeing \( f(n) \) increments by 0.5, note after k steps: \( f(n+k) = f(n) + 0.5k \). Since we start with \( f(1) = 2 \), \( f(101) = 2 + 0.5(100) \).

Step 5: Solve for f(101)

Using the pattern \( f(101) = 2 + 0.5 \times 100 = 2 + 50 = 52 \). Therefore, \( f(101) = 52 \).

Key Concepts

Mathematical Induction

Mathematical induction is an essential proof technique in mathematics. It is commonly used for proving statements about integers, particularly sequences or series. The process involves two main steps:

• Base Case: First, show that the statement holds for an initial value, usually the smallest integer that meets the conditions of the problem, like n=1.
• Inductive Step: Prove that if the statement holds for some arbitrary integer k, then it also holds for k+1.

By following these steps, you can establish that the statement is true for all integers greater than or equal to the initial value.

In the context of recurrence relations like the one in our problem, induction can be used to prove patterns or formulas derived from such relations. Once a pattern is suspected - like the increment by 0.5 discovered here - induction helps to rigorously establish its truth for every step in the sequence.

Sequences and Series

Sequences and series are fundamental concepts in mathematics that deal with ordered lists of numbers and their summation, respectively. A sequence is essentially a list of numbers in a specific order, while a series is the sum of terms of a sequence.

• In our exercise, we are dealing with a numerical sequence defined by a recurrence relation, where each term is defined based on the previous term.
• Recognizing the type of sequence, whether arithmetic, geometric, or something else, is crucial as it dictates how the sequence progresses.

In this case, we have identified a pattern in the sequence: each subsequent term increases by 0.5. This regular increase is characteristic of an arithmetic sequence, where the difference between consecutive terms is constant. Understanding this helps predict future terms without recalculating each one, speeding up problem-solving significantly.

Pattern Recognition

Pattern recognition is a critical skill in solving complex problems, especially when working with sequences or recursive formulas. It involves identifying regularities, making observations, and predicting future occurrences based on those observations.

• When given a few terms, focus on how each term relates to the previous one. Are they increasing or decreasing? By how much?
• In our situation, by computing the first few terms \(f(1) = 2\), \(f(2) = 2.5\), \(f(3) = 3\), and so on, it becomes clear there's a pattern of increment by 0.5 per step.
• Extrapolate this pattern: once confirmed, use it to predict terms beyond those already computed, like \(f(101)\) in this case.

Efficiently recognizing patterns can transform a complicated recurrence problem into a simple arithmetic calculation, saving time and minimizing errors.