Question

If A and B are the coefficients of $x^n$ in the expansion of $(1 + x)^2n$ and $(1 + x)^2n - 1$, then:

• A. $A = B$
• B. $2A = B$
• C. $A = 2B$
• D. $AB = 2$

Hint:

Remember binomial symmetry: $\binomnr = \binomnn-r$ and identities like $\binomnr = \fracnr \binomn - 1r - 1$ for quick conversions.

Answer 1

The Correct Option is C

Step 1: The coefficient of $x^n$ in $(1 + x)^2n$ is given by the binomial coefficient: \[ A = \binom2nn \] Step 2: The coefficient of $x^n$ in $(1 + x)^2n - 1$ is given by: \[ B = \binom2n - 1n \] Step 3: Use the binomial coefficient identity: \[ \binom2nn = \frac2nn \binom2n - 1n - 1 = 2 \binom2n - 1n - 1 \] Step 4: Relate $\binom2n - 1n$ and $\binom2n - 1n - 1$: From the symmetry property: \[ \binom2n - 1n - 1 = \binom2n - 1n \] Therefore: \[ A = 2 \binom2n - 1n = 2B \] Step 5: Conclusion: \[ A = 2B \] Thus, the correct answer is $\mathbfA = 2B$.