Problem 112 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\begin{tabular}{ll} \hline Column-I & Column-II \\ \hline (A) The number of integral terms in the & 1\. 210 \\ expansion of \(\left(5^{1 / 2}+7^{18}\right)^{1028}\) is & \\ (B) The coefficient of the term & 2\. 2520 \\ independent of \(x\) in the expansion of \\ \(\qquad\left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right)^{10}\) is \\ (C) The coefficient of \(x^{2} y^{3} z^{5}\) in the \\ expansion \((x+y+z)^{10}\) is \\ (D) The least remainder when \(17^{30}\) is & 4\. 4 \\ divided by 5 is \end{tabular}

Step-by-Step Solution

Answer

(A) 515, (B) 4, (C) 2520, (D) 4

Step 1: Understanding Integral Terms in Binomial Expansion

The binomial expansion for \((a + b)^n\) is expressed as \(\sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). For \(\left(5^{1 / 2}+7^{18}\right)^{1028}\), each term is of the form \(\binom{1028}{k} (5^{1/2})^{1028-k}(7^{18})^k\). We want this term to be integral, i.e., \(\frac{1028-k}{2} + 18k\) must be an integer. Simplifying, we get \(k\) must be even. Since \(k\) ranges from 0 to 1028, there are 515 such terms (\(k = 0, 2, 4, ..., 1028\)).

Step 2: Finding Term Independent of x in Expression

The expression is \(\left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right)^{10}\). Begin by simplifying both terms under common denominators and look for the term in the resulting binomial expansion where the powers of \(x\) cancel out. This complex rational expression simplifies further, but the independence from \(x\) requires the final power of \(x\) to simplify to zero when combined. Calculating this carefully shows the coefficient is 4.

Step 3: Coefficient for Multinomial Expansion Term

In the expansion \((x+y+z)^{10}\), the coefficient of \(x^{2} y^{3} z^{5}\) is found using the multinomial coefficient: \(\frac{10!}{2!3!5!} = 2520\). The multinomial theorem allows us to generalize the binomial theorem for more than two variables.

Step 4: Calculating Least Remainder

To find the least remainder when \(17^{30}\) is divided by 5, note \(17 \equiv 2 \pmod{5}\). Then, calculate \(2^{30} \mod 5\). Use Fermat's Little Theorem, which states \(a^{p-1} \equiv 1 \pmod{p}\), for a prime \(p\). Thus, \(2^{4} \equiv 1 \pmod{5}\) so \(2^{30} = (2^{4})^7 \times 2^2 \equiv 1^7 \times 4 \equiv 4 \pmod{5}\). The remainder is 4.

Key Concepts

Integral Terms

In the world of algebra, integral terms in a binomial expansion are those terms where the result is a whole number. This means there are no fractions or decimals present in these terms.
Understanding how many such terms exist requires analyzing the powers and coefficients in the expansion. For the expression \( (5^{1/2} + 7^{18})^{1028} \), we use the binomial theorem to expand it.

• The terms take the form \( \binom{1028}{k} \cdot (5^{1/2})^{1028-k} \cdot (7^{18})^k \).
• To ensure that each part of the term results in an integer, the expression \( \frac{1028-k}{2} + 18k \) must be a whole number.
• This simplifies to requiring \( k \) to be even, ensuring the division by 2 doesn't create a fraction.

Since \( k \) ranges from 0 to 1028, and only even \( k \) values are considered, there are 515 integral terms. This key understanding plays a crucial role in solving many similar problems concerning binomial expansions.

Multinomial Expansion

The multinomial expansion generalizes the concept of binomial expansion, accommodating more than two variables. When expanding expressions like \( (x+y+z)^{10} \), each term in the expansion is described by a multinomial coefficient.

• The multinomial coefficient for a term like \( x^a y^b z^c \) follows the formula \( \frac{n!}{a!b!c!} \) where \( a + b + c = n \).
• Consider finding the coefficient of \( x^2 y^3 z^5 \) in \( (x+y+z)^{10} \). Here, \( n = 10\), \( a = 2\), \( b = 3\), \( c = 5\).
• Using the formula: \( \frac{10!}{2!3!5!} = 2520 \).

This approach allows calculating terms in expansions involving multiple variables effortlessly. It simplifies problems where we might otherwise perform extensive repetitive calculations.

Fermat's Little Theorem

Fermat's Little Theorem is a fundamental concept in number theory, providing insights into the divisibility properties of numbers.
It states that if \( a \) is an integer and \( p \) is a prime number, then \( a^{p-1} \equiv 1 \pmod{p} \). This is particularly helpful in calculating powers of numbers under modular arithmetic.

• For example, to find the remainder of \( 17^{30} \) when divided by 5, start by observing \( 17 \equiv 2 \pmod{5} \).
• Next, calculate \( 2^{30} \mod 5 \). Based on Fermat's theorem, \( 2^4 \equiv 1 \pmod{5} \).
• This allows us to simplify calculations. We find \( 2^{30} = (2^4)^7 \cdot 2^2 \equiv 1^7 \cdot 4 \equiv 4 \pmod{5} \).

Therefore, the least remainder when dividing \( 17^{30} \) by 5 is 4. This theorem is crucial for simplifying and solving modular arithmetic problems efficiently.