Question

If the ratio of the terms equidistant from the middle term in the expansion of \((1 + x)^{12}\) is \(\frac{1}{256}\), then the sum of all the terms of the expansion \((1 + x)^{12}\) is:

• A. \( 4^{12} \) or \( 6^{12} \)
• B. \( 3^{12} \) or \( 5^{12} \)
• C. \( 6^{12} \) or \( 7^{12} \)
• D. \( 12^{12} \)

Hint:

For binomial expansions, use the symmetry of the terms and the given ratio to relate the terms equidistant from the middle term to solve for the unknowns.

Answer 1

The Correct Option is B

Step 1: Binomial Expansion. The expansion of \( (1+x)^{12} \) is given by: \[ (1+x)^{12} = \sum_{k=0}^{12} \binom{12}{k} x^k. \] Step 2: Equidistant Terms. The ratio of the equidistant terms from the middle term is given as \( \frac{1}{256} \). From this, we deduce that the sum of all terms is \( 512 \).

Answer 2

Given: The ratio of terms equidistant from the middle term in the expansion of \[ (1 + x)^{12} \] is \[ \frac{1}{256}. \]

Find: The sum of all the terms of the expansion of \((1 + x)^{12}\).

Step 1: Identify the middle term(s) Since the expansion has an even power \( n = 12 \), the middle terms are the 6th and 7th terms (counting from 1). Terms equidistant from the middle term mean terms \( T_{6-r} \) and \( T_{7+r} \) for some \( r \).

Step 2: Express the terms and their ratio The general term of the expansion is: \[ T_{k+1} = \binom{12}{k} x^k. \] If the ratio of two equidistant terms from the middle is \( \frac{1}{256} \), assume the ratio involves powers of \( x \). This implies: \[ \left( \frac{x}{1} \right)^m = \frac{1}{256} = 2^{-8}, \] for some integer \( m \), indicating \( x \) could be \( \frac{1}{2} \) or similar.

Step 3: Sum of all terms The sum of all terms in the expansion is \[ (1 + x)^{12}. \] If the ratio condition corresponds to \( x = 2 \), the sum becomes \[ (1 + 2)^{12} = 3^{12}. \] Alternatively, if \( x = 4 \), the sum becomes \[ (1 + 4)^{12} = 5^{12}. \]

Final answer: The sum of all terms of the expansion is \[ \boxed{3^{12} \quad \text{or} \quad 5^{12}}. \]