Question

Sum of the last 40 coefficients in the expansion of $ (1 + x)^{79} $, when expanded in ascending power of $ x $ is:

• A. \( 2^{79} \)
• B. \( 2^{40} \)
• C. \( 2^{39} \)
• D. \( 2^{78} \)

Hint:

To find the sum of coefficients in the expansion of \( (1 + x)^n \), evaluate the expansion at \( x = 1 \) and subtract the sum of the first few coefficients if needed.

Answer 1

The Correct Option is B

The binomial expansion of \( (1 + x)^{79} \) is given by:
\[ (1 + x)^{79} = \sum_{k=0}^{79} \binom{79}{k} x^k \] The coefficients of the terms are the binomial coefficients \( \binom{79}{k} \). 
Step 1:
The sum of the coefficients of the terms \( x^k \) for \( k = 40, 41, \dots, 79 \) is required. 
Step 2:
The sum of the coefficients of the expansion is the value of the expression when \( x = 1 \): \[ (1 + 1)^{79} = 2^{79} \] 
Step 3:
The sum of the first 40 coefficients is the sum for \( k = 0 \) to \( k = 39 \), and the sum of the last 40 coefficients is 
given by:
\[ \text{Sum of last 40 coefficients} = 2^{79} - \text{Sum of first 39 coefficients} \] Using symmetry in the binomial coefficients, the sum of the last 40 coefficients is equal to \( 2^{40} \).