Question

If a and b are the greatest values of \( {}^{2n}C_{r} \) and \( {}^{(2n-1)}C_{r} \) respectively, then

• A. \( a=2b \)
• B. \( b=2a \)
• C. \( a=b \)
• D. \( a^{2}=2b^{2} \)

Hint:

Test with a simple value like \( n=2 \): \[ a = \max({}^4C_r) = {}^4C_2 = 6 \] \[ b = \max({}^3C_r) = {}^3C_1 = 3 \] Since \( 6 = 2(3) \), the relation \( a = 2b \) is verified instantly!

Answer 1

The Correct Option is A

Concept: For any binomial coefficient \( {}^{m}C_{r} \), the maximum value occurs at the middle term:

• If \( m \) is even, the greatest value occurs at \( r = \frac{m}{2} \).

• If \( m \) is odd, the greatest values occur at \( r = \frac{m-1}{2} \) and \( r = \frac{m+1}{2} \).

Step 1: Finding the greatest value \( a \).
The upper index is \( 2n \), which is an even number. Therefore, the greatest value \( a \) occurs at \( r = \frac{2n}{2} = n \): \[ a = {}^{2n}C_{n} \]

Step 2: Finding the greatest value \( b \).
The upper index is \( 2n-1 \), which is an odd number. Therefore, the greatest value \( b \) occurs at \( r = \frac{(2n-1)-1}{2} = n-1 \): \[ b = {}^{2n-1}C_{n-1} \]

Step 3: Establishing the relationship between \( a \) and \( b \).
Let us expand \( a = {}^{2n}C_{n} \) using the standard factorial definition: \[ a = \frac{(2n)!}{n! \cdot n!} = \frac{2n \cdot (2n-1)!}{n \cdot (n-1)! \cdot n!} = \frac{2n}{n} \cdot \frac{(2n-1)!}{(n-1)! \cdot n!} \] \[ a = 2 \cdot \frac{(2n-1)!}{(n-1)! \cdot [(2n-1) - (n-1)]!} = 2 \cdot {}^{2n-1}C_{n-1} \] Substituting \( b = {}^{2n-1}C_{n-1} \): \[ a = 2b \]