Problem 83 - A Complete Resource Book in Mathematics for JEE Main

Question

For all \(n \in N\), the integer just above \((\sqrt{3}+1)^{2 n}\) is divisible by (A) \(2^{\mathrm{n}+1}\) (B) \(2^{\mathrm{n}}+1\) (C) \(2^{\mathrm{n}+1}+1\) (D) none of these

Step-by-Step Solution

Answer

Option (A): \(2^{n+1}\) is the correct answer.

Step 1: Understand the Problem

We need to find which option correctly states the divisibility of the integer just above \((\sqrt{3}+1)^{2n}\). We do this by evaluating the expression \((\sqrt{3}+1)^{2n}\) and finding the smallest integer greater than this value.

Step 2: Rewrite the Expression

Express \((\sqrt{3}+1)^{2n}\) as a binomial expansion: \[(\sqrt{3}+1)^{2n} = a_n + b_n \sqrt{3}\] where \(a_n\) and \(b_n\) are integers obtained from the binomial expansion.

Step 3: Use Conjugate Properties

To simplify, examine the expression \((\sqrt{3}-1)^{2n} = a_n - b_n \sqrt{3}\). Both expansions together help isolate integer parts and allow understanding of how terms involving \(b_n\sqrt{3}\) cancel out.

Step 4: Sum Conjugate Results

Adding the two conjugate expressions results in\[ (\sqrt{3}+1)^{2n} + (\sqrt{3}-1)^{2n} = 2a_n \]This implies \(2a_n\) is an integer and is the sum of real numbers from both expanded forms.

Step 5: Determine The Integer Above

The integer just above \((\sqrt{3}+1)^{2n}\) is \(a_n + 1\) since the fractional part \(b_n \sqrt{3}\) is less than 1 when \((\sqrt{3}-1)^{2n}\) is very small for large \(n\).

Step 6: Check Divisibility for n

For \(n=0\), check the options. Simplifying gives the integer above as \(2\). Option (A) \(2^{n+1}\) matches as \(2^{0+1} = 2\) and is divisible by 2.

Step 7: Validate General Case

In general for several values of \(n\), evaluate the expression using recursion, or algebraic identities, showing the pattern that the integer part is dependent on powers of 2.

Step 8: Choose Correct Option

Re-check calculations for \((\sqrt{3}+1)^{2n}\)'s integer approximation consistency with \(2^{n+1}\) divisibility.

Key Concepts

Integer Divisibility

Integer divisibility is a fundamental concept in number theory. It concerns whether one integer can be divided by another without leaving a remainder. This is often denoted by saying that the integer is 'divisible by' another.

• Key Idea: An integer \(a\) is divisible by an integer \(b\) if there exists an integer \(k\) such that \(a = b \times k\).
• To solve divisibility problems, like the one in this exercise, often requires checking modularity or using known divisibility rules.

In our exercise, we are determining whether the integer just above \((\sqrt{3}+1)^{2n}\) is divisible by a specific power of 2. This requires understanding how to round the result of an expression to an integer and check the divisibility of that integer. This divisibility check relies on examining the resulting integer's factors and simplifying the expression to fit into a divisibility rule, in this case for powers of 2.

Binomial Theorem

The Binomial Theorem provides a way to expand expressions that are raised to a power. It is particularly useful in algebra whenever you encounter expressions like \((a+b)^n\).

• Formula: The theorem states that \((a+b)^n = \sum_{k=0}^{n} {\binom{n}{k} a^{n-k} b^k}\), where \(\binom{n}{k}\) is a binomial coefficient.
• Binomial expansions result in terms that can be organized into integer and non-integer parts, which is key in this exercise.
• Usage: For \((\sqrt{3}+1)^{2n}\), we break the expression into \(a_n + b_n \sqrt{3}\) by identifying integer and irrational parts.

Using the binomial theorem in this problem simplifies the calculation of high powers, making them more manageable, and helps us separate integer parts from expressions that contain square roots.

Algebraic Identities

Algebraic identities are equations that hold for all values of the variables involved. They allow us to manipulate and simplify expressions efficiently, an invaluable skill in mathematics.

• Example: \((a+b)(a-b) = a^2 - b^2\) is a basic algebraic identity utilized to separate terms in binomial expressions.
• Application: In our problem, observing the conjugate \((\sqrt{3}-1)^{2n}\) and combining it with \((\sqrt{3}+1)^{2n}\) helps eliminate the irrational part of \(b_n \sqrt{3}\).
• This method highlights the power of identities in simplifying complex roots and powers just by using real algebraic manipulations.

The conjugate technique utilized in our exercise leverages algebraic identities to transform a complex expression into a simpler integer-focused equation with clearer divisibility, ultimately pointing us towards the correct option.