Understanding Binomial Expansion Coefficients

The question asks us to find the value of 'r' for which the coefficients of the terms containing \(x^r\) and \(x^{r+1}\) are equal in the binomial expansion of \((1 + x)^{2n + 1}\).

General Term in Binomial Expansion

The binomial theorem states that the expansion of \((a+b)^N\) is given by:

\((a+b)^N = \sum_{k=0}^{N} \binom{N}{k} a^{N-k} b^k\)

The general term (the \((k+1)^{th}\) term) in this expansion is \(T_{k+1} = \binom{N}{k} a^{N-k} b^k\). The coefficient of \(b^k\) is \(\binom{N}{k}\).

In our case, we have the expansion of \((1+x)^{2n+1}\). Here, \(a=1\), \(b=x\), and \(N=2n+1\). The general term is:

\(T_{k+1} = \binom{2n+1}{k} (1)^{(2n+1)-k} (x)^k = \binom{2n+1}{k} x^k\)

The coefficient of \(x^k\) in the expansion of \((1+x)^{2n+1}\) is \(\binom{2n+1}{k}\).

Identifying Coefficients of \(x^r\) and \(x^{r+1}\)

Based on the general term, we can identify the coefficients:

• The coefficient of \(x^r\) corresponds to \(k=r\). So, the coefficient of \(x^r\) is \(\binom{2n+1}{r}\).
• The coefficient of \(x^{r+1}\) corresponds to \(k=r+1\). So, the coefficient of \(x^{r+1}\) is \(\binom{2n+1}{r+1}\).

Setting the Coefficients Equal

According to the problem statement, the coefficient of \(x^r\) and the coefficient of \(x^{r+1}\) are equal. Therefore, we can write the equation:

\(\binom{2n+1}{r} = \binom{2n+1}{r+1}\)

Solving the Equation using Binomial Coefficient Property

We use a fundamental property of binomial coefficients: If \(\binom{N}{k} = \binom{N}{m}\), where \(0 \le k, m \le N\), then either \(k = m\) or \(k + m = N\).

In our equation \(\binom{2n+1}{r} = \binom{2n+1}{r+1}\), we have \(N=2n+1\), \(k=r\), and \(m=r+1\). Applying the property:

Case 1: \(k = m\)

\(r = r+1\)

Subtracting 'r' from both sides gives \(0 = 1\), which is impossible.

Case 2: \(k + m = N\)

\(r + (r+1) = 2n+1\)

Simplify the left side:

\(2r + 1 = 2n+1\)

Subtract 1 from both sides:

\(2r = 2n\)

Divide both sides by 2:

\(r = n\)

This gives us the value of 'r' for which the coefficients are equal.

Verification

If \(r=n\), the coefficients are \(\binom{2n+1}{n}\) and \(\binom{2n+1}{n+1}\). We know that \(\binom{N}{k} = \binom{N}{N-k}\). So, \(\binom{2n+1}{n+1} = \binom{2n+1}{(2n+1)-(n+1)} = \binom{2n+1}{2n+1-n-1} = \binom{2n+1}{n}\). This confirms that when \(r=n\), the coefficients are indeed equal.

Conclusion

The value of \(r\) for which the coefficients of \(x^r\) and \(x^{r+1}\) are equal in the expansion of \((1 + x)^{2n + 1}\) is \(n\).

Revision Table: Key Concepts

Additional Information: Binomial Expansion Properties

The binomial expansion \((1+x)^N\) has several interesting properties related to its coefficients:

• Number of terms: The expansion of \((1+x)^N\) has \(N+1\) terms.
• Sum of coefficients: If we set \(x=1\) in the expansion, the sum of all coefficients is \((1+1)^N = 2^N\).
• Alternating sum of coefficients: If we set \(x=-1\) in the expansion, the alternating sum of coefficients is \((1-1)^N = 0^N\) (which is 0 for \(N > 0\) and 1 for \(N=0\)).
• Maximum coefficient: For even \(N\), the largest coefficient is the middle term, \(\binom{N}{N/2}\). For odd \(N\), there are two equal largest coefficients in the middle, \(\binom{N}{(N-1)/2}\) and \(\binom{N}{(N+1)/2}\). This question relates directly to finding the conditions for two adjacent coefficients to be equal, which happens at the peak(s) of the coefficient distribution.