Question

If the coefficients of \((5r+4)th\) term and \((r-1)th\) term in the expansion of \((1+x)^{25}\) are equal, then \(r\) is 

• A. \(6\)
• B. \(3\)
• C. \(5\)3
• D. \(2\)
• E. \(4\)

Answer 1

Answer: (E)

Given: In the expansion of \( (1+x)^{25} \), the coefficients of the \( (5r+4)^{th} \) term and \( (r-1)^{th} \) term are equal.

We know the general term in the binomial expansion is: \[ T_{k+1} = \binom{25}{k}x^k \]

Therefore: 1. The \( (5r+4)^{th} \) term corresponds to \( k = 5r+3 \) 2. The \( (r-1)^{th} \) term corresponds to \( k = r-2 \)

Setting their coefficients equal: \[ \binom{25}{5r+3} = \binom{25}{r-2} \]

This equality holds when: 1. \( 5r+3 = r-2 \) (which gives \( r = -\frac{5}{4} \), invalid as \( r \) must be positive integer), or 2. \( 5r+3 + (r-2) = 25 \) (by the symmetry property of binomial coefficients)

Solving the second case: \[ 6r + 1 = 25 \] \[ 6r = 24 \] \[ r = 4 \]

The correct answer is (E) 4.

Answer 2

Given

\((5r + 4)\text{th  term} = 25C_{5r+3}x^{5r+4}\)

\((r − 1)^{1/4}\text{  term } = 25C_{r−2}x^ {r−1}\)

\(25C_{5r+3} x^{5r+4} = 25C_{r−2}\)

\((5r + 3) + (r − 2) = 25\)

\(6r+1=25\)

\(r=4\)