Question

Sum of coefficients of the last 6 terms in the expansion of \( (1 + x)^{11} \) when the expansion is in ascending powers of \( x \), is

• A. \(2048\)
• B. \(32\)
• C. \(512\)
• D. \(64\)
• E. \(1024\)

Hint:

For binomial expansions, sum of all coefficients is \(2^n\). Use symmetry to quickly find partial sums.

Answer 1

Answer: (E)

Concept: The binomial expansion: \[ (1+x)^{11} = \sum_{k=0}^{11} \binom{11}{k} x^k \] Thus, coefficients are: \[ \binom{11}{0}, \binom{11}{1}, \ldots, \binom{11}{11} \]

Step 1: Identify last 6 terms
Total terms = 12 Last 6 terms correspond to: \[ k = 6,7,8,9,10,11 \] Thus required sum: \[ \binom{11}{6} + \binom{11}{7} + \binom{11}{8} + \binom{11}{9} + \binom{11}{10} + \binom{11}{11} \]

Step 2: Use symmetry property
\[ \binom{11}{k} = \binom{11}{11-k} \] So: \[ \binom{11}{6} = \binom{11}{5}, \quad \binom{11}{7} = \binom{11}{4}, \ldots \] Thus last 6 terms = first 6 terms.

Step 3: Sum of all coefficients
\[ (1+1)^{11} = 2^{11} = 2048 \]

Step 4: Half sum
Since coefficients are symmetric: \[ \text{Sum of first 6 terms} = \text{Sum of last 6 terms} = \frac{2048}{2} = 1024 \] Final Answer: \[ \boxed{1024} \]