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

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

The number of integral terms is 107.

Step 1: Understand the Binomial Expansion

The expression \((2\sqrt{5} + \sqrt[6]{7})^{642}\) can be expanded using the Binomial Theorem, which states: \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). Here, \(a = 2\sqrt{5}\), \(b = \sqrt[6]{7}\), and \(n = 642\).

Step 2: Determine the General Term

The general term in the expansion is given by \(T_k = \binom{642}{k} (2\sqrt{5})^{642-k} (\sqrt[6]{7})^k\). Simplifying this, \(T_k = \binom{642}{k} 2^{642-k} 5^{(642-k)/2} 7^{k/6}\).

Step 3: Condition for Integral Terms

For the term \(T_k\) to be an integer, both the powers of 5 and 7 must be integers. Thus, \((642-k)/2\) and \(k/6\) must be integers. This translates to two conditions: 1. \(642-k\) must be even. 2. \(k\) must be a multiple of 6.

Step 4: Solve the Condition for Even Power of 2

Since \(642\) is even, \((642-k)\) also needs to be even, which means \(k\) should be even. This condition will naturally be satisfied if \(k\) is a multiple of 6.

Step 5: Solve the Condition for Multiple of 6

Find the range of \(k\) values that are multiples of 6. Since \(k\) ranges from 0 to 642, we calculate the smallest and largest multiples of 6 within this range: \(k_{min} = 0\) and \(k_{max} = 642\). So, the multiples of 6 within this range are \(0, 6, 12, \ldots, 642\).

Step 6: Calculate Number of Valid k Values

Count the number of multiples of 6 from \(0\) to \(642\). This is an arithmetic sequence where the first term \(a = 0\), last term \(l = 642\), and the common difference \(d = 6\). The number of terms \(n\) is calculated by the formula \(n = \frac{l-a}{d} + 1\). Thus, \(n = \frac{642-0}{6} + 1 = 107\).

Step 7: Conclude the Solution

The number of integral terms is the same as the number of valid \(k\) values, which is 107.

Key Concepts

Integral Terms

In mathematics, when we talk about integral terms in a binomial expansion, we refer to terms that have integer values. For a term to be considered an integral term in a binomial expansion, all parts of the term must result in whole numbers.
In our exercise, the expression is given by \((2\sqrt{5}+\sqrt[6]{7})^{642}\). Each term in the expansion is of the form \(T_k = \binom{642}{k} (2\sqrt{5})^{642-k} (\sqrt[6]{7})^k\).
For \(T_k\) to be an integer, both the exponents of the numbers must lead to integer results.

• This means that \((642-k)/2\) must be an integer so that the power of 5 remains integral.
• Similarly, \(k/6\) must be an integer making sure that the power of 7 remains integral as well.

Therefore, the conditions necessary are that \(642-k\) must be even, and \(k\) a multiple of 6.

Binomial Expansion

The binomial theorem provides a way to expand expressions of the form \((a+b)^n\), where \(a\) and \(b\) are any numbers, and \(n\) is a non-negative integer. The theorem is stated as: \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\).
This expansion results in a series of terms, where each term is generated based on different combinations of powers of \(a\) and \(b\).
For example, for the expression \((2\sqrt{5} + \sqrt[6]{7})^{642}\), the expansion involves terms such as \(\binom{642}{k} (2\sqrt{5})^{642-k} (\sqrt[6]{7})^k\). Here, \(\binom{642}{k}\) is the binomial coefficient which gives the number of ways to choose \(k\) elements from a set of \(642\) elements and is integral to determining the final value of each term in the expansion.

Arithmetic Sequence

An arithmetic sequence is a sequence of numbers in which each term after the first is generated by adding a constant, known as the common difference, to the previous term. This is a straightforward, yet powerful concept that often appears in problems related to binomial expansions.
In the exercise, determining the number of integral terms involves finding those \(k\) values which satisfy the condition \(k/6\). These \(k\) values form an arithmetic sequence starting from \(0\) and ending at \(642\), with a common difference of \(6\).

• The first term \(a = 0\) and the last term \(l = 642\).
• The sequence includes all multiples of 6 from \(0\) to \(642\).

To find the number of terms in this sequence, use the formula \(n = \frac{l-a}{d} + 1\). In this instance, \(n = \frac{642-0}{6} + 1 = 107\). Thus, there are 107 integral values for \(k\) that satisfy both conditions, confirming 107 integral terms in the expansion.