Problem 19 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The number of integral terms in the expansion of \((2 \sqrt{5}+\sqrt[6]{7})^{642}\) is (A) 105 (B) 107 (C) 321 (D) 108

Step-by-Step Solution

Answer

108

Step 1: Understanding the problem

In the expansion of \((2 \sqrt{5} + \sqrt[6]{7})^{642}\), we need to find which terms result in integers. In the binomial expansion, the general term is calculated, and we must check when this term is an integer.

Step 2: General Term of the Expansion

The general term of the expansion is given by \(T_k = \binom{642}{k} (2 \sqrt{5})^{642-k} (\sqrt[6]{7})^k\). We need to find the values of \(k\) for which \(T_k\) is an integer.

Step 3: Analyzing the Term for Integrality

The term becomes integral if the powers of \(\sqrt{5}\) and \(\sqrt[6]{7}\) end up being integers. That means:\((2 \sqrt{5})^{642-k} = 2^{642-k} \cdot 5^{(642-k)/2}\) must be a perfect square, and \(\sqrt[6]{7})^k = 7^{k/6}\) must be an integer.

Step 4: Condition for Perfect Square

For \(5^{(642-k)/2}\) to be an integer, \(\frac{642-k}{2}\) must be an integer. This implies \(642 - k\) must be even, so \(k\) needs to be even.

Step 5: Condition for Integer Powers of 7

For \(7^{k/6}\) to be an integer, \(\frac{k}{6}\) needs to be an integer. Therefore, \(k\) must be a multiple of 6.

Step 6: Finding the Range of k

Since \(k\) must be both even and a multiple of 6, it follows \(k\) must be the least common multiple of 6 and 2, which is 6. Therefore, \(k\) must be a multiple of 6. The maximum value for \(k\) is 642.

Step 7: Counting the Valid k Values

The sequence of \(k\) values being multiples of 6 starting from 0 up to 642 is: 0, 6, 12, ..., 642. This forms an arithmetic sequence with the first term as 0, the common difference as 6, and the last term as 642.

Step 8: Arithmetic Sequence Calculation

The number of terms in the sequence is calculated by the formula for the number of terms in an arithmetic sequence: \(n = \frac{l-a}{d} + 1\), where \(l\) is the last term, \(a\) is the first term, and \(d\) is the common difference. Substituting gives \(n = \frac{642 - 0}{6} + 1 = 108\).

Key Concepts

Integral Terms

Integral terms in a binomial expansion are those where the resulting expression is a whole number, with no fractional or decimal component. To identify integral terms within the expansion of \((2 \sqrt{5} + \sqrt[6]{7})^{642}\), we must ensure that each component within the term is adjusted to make the entire expression an integer.

• We look at the powers involved: \((2 \sqrt{5})^{642-k}\) and \((\sqrt[6]{7})^k\). To achieve integer results, the exponents need to satisfy specific criteria.
• For \((2 \sqrt{5})^{642-k}\), the exponent must allow it to resolve into a perfect square.
• For \((\sqrt[6]{7})^k\), the exponent must elevate to a power that produces an integer.

Through careful observation, it becomes evident that solving these constraints allows us to determine which terms contribute entire, integral values to the expansion's result.

Arithmetic Sequence

An arithmetic sequence is a series of numbers in which the difference between consecutive terms remains consistent. In the context of our problem, the values of \(k\) form such a sequence because each term in this series is derived by incrementing its predecessor by a fixed amount, known as the common difference. For the expansion of \((2 \sqrt{5} + \sqrt[6]{7})^{642}\), determining which \(k\) values lead to integral terms involves identifying a suitable arithmetic sequence.

• Here, we know \(k\) must be a multiple of 6 due to the requirements for integrality we explored earlier.
• This criterion results in \(k\) values like 0, 6, 12, ..., up to 642, maintaining a common difference of 6.
• Conclusively, we have an arithmetic sequence defined by its first term (0), common difference (6), and last term (642).

This foundational tool allows us to calculate how many valid \(k\) terms exist, essential for our problem's solution.

Perfect Square

A perfect square refers to a number that can be expressed as the square of an integer. The existence of perfect squares is pivotal in ensuring that components like \((2 \sqrt{5})^{642-k}\) generate integral outcomes within the expansion.

• For a term involving \(2^{642-k} \cdot 5^{(642-k)/2}\), ensuring it's a perfect square requires that \((642-k)/2\) is an integer.
• This condition ensures \(5^{n/2}\) (with \(n = 642-k\)) does not yield fractional components, thereby maintaining integrality throughout the calculation.

Understanding how perfect squares function aids in ensuring the integrality of terms, propelling the sequence of calculations needed in solving problems involving large expansions.

Exponentiation

Exponentiation involves raising a base number to the power of an exponent, resulting in repetitive multiplication. In our problem, exponentiation applies to both \((2 \sqrt{5})\) and \((\sqrt[6]{7})\), playing a critical role in determining the integrality of terms in their expanded form.

• The concept engages by raising these bases to powers designated by expressions like \((642-k)\) and \(k\).
• For integrality, it becomes crucial that these exponents result in integer values when applied to respective bases.
• This requires careful selection of \(k\) to satisfy principles like base \((\sqrt[6]{7})\) being raised to a power divisible by 6 for integer results.

Through understanding exponentiation, we achieve a clearer view of the operations and considerations necessary to solve for integer outcomes within complex binomial expansions.