Understanding the Sum of Coefficients in Binomial Expansions

The question asks for the sum of the coefficients of all the terms in the expansion of the binomial expression $(1 + x)^{2n + 1}$.

A fundamental property of binomial expansions is that the sum of the coefficients of all terms in the expansion of $(a + bx)^N$ can be found by simply substituting the variable (in this case, $x$) with 1. This is because when the variable is 1, each term in the expansion reduces to its coefficient.

Let the expansion of $(1 + x)^{2n + 1}$ be represented by:

\((1 + x)^{2n + 1} = c_0 x^0 + c_1 x^1 + c_2 x^2 + \dots + c_{2n+1} x^{2n+1}\)

Here, \(c_0, c_1, \dots, c_{2n+1}\) are the coefficients of the terms in the expansion.

To find the sum of these coefficients, \(c_0 + c_1 + \dots + c_{2n+1}\), we substitute \(x = 1\) into the expansion:

\((1 + 1)^{2n + 1} = c_0 (1)^0 + c_1 (1)^1 + c_2 (1)^2 + \dots + c_{2n+1} (1)^{2n+1}\)

\((2)^{2n + 1} = c_0 (1) + c_1 (1) + c_2 (1) + \dots + c_{2n+1} (1)\)

\(2^{2n + 1} = c_0 + c_1 + c_2 + \dots + c_{2n+1}\)

So, the sum of the coefficients of all the terms in the expansion of $(1 + x)^{2n + 1}$ is \(2^{2n + 1}\).

Comparing the Result with Given Options

Now, let's examine the given options and see which one matches our calculated sum, \(2^{2n + 1}\):

1. <p>2 <sup>2n-1</sup></p> corresponds to \(2^{2n-1}\). This is not \(2^{2n+1}\).
2. <p>4 <sup>n-1</sup></p> corresponds to \(4^{n-1}\). Since \(4 = 2^2\), this is \((2^2)^{n-1} = 2^{2(n-1)} = 2^{2n - 2}\). This is not \(2^{2n+1}\).
3. <p>2 &times; 4 <sup>n</sup></p> corresponds to \(2 \times 4^n\). Since \(4 = 2^2\), this is \(2 \times (2^2)^n = 2 \times 2^{2n}\). Using the rule of exponents \(a^m \times a^n = a^{m+n}\), this becomes \(2^1 \times 2^{2n} = 2^{1 + 2n}\). This matches our calculated sum \(2^{2n + 1}\).
4. <p>None of the above</p> - Since Option 3 matches, this option is incorrect.

Therefore, the sum of the coefficients of all the terms in the expansion of \((1 + x)^{2n + 1}\) is equal to \(2 \times 4^n\).

Revision Table: Binomial Expansion Coefficients

Additional Information on Binomial Expansion

The sum of coefficients is a useful property often tested. For a general binomial expansion \((a+bx)^N\), the sum of coefficients is found by substituting \(x=1\), resulting in \((a+b \times 1)^N = (a+b)^N\). In the specific case of \((1+x)^{2n+1}\), \(a=1\), \(b=1\), and \(N=2n+1\), leading to the sum of coefficients being \((1+1)^{2n+1} = 2^{2n+1}\).

Understanding the structure of binomial coefficients \(\binom{N}{k}\) is also important. These are often represented by Pascal's Triangle, where each number is the sum of the two numbers directly above it. The sum of the numbers in the N-th row of Pascal's Triangle (corresponding to the coefficients of \((x+y)^N\)) is always \(2^N\).

For the expansion of \((1-x)^N\), the sum of coefficients is \((1-1)^N = 0^N\). If \(N \ge 1\), the sum is 0. If \(N=0\), \((1-x)^0 = 1\) (for \(x \neq 1\)), and the sum of coefficients is 1. This shows how the sign of the variable affects the sum of coefficients.