Question

Sum of the last 30 coefficients in the expansion of \[ (1 + x)^{59}, \] when expanded in ascending powers of \( x \), is:

• A. \(2^{59}\)
• B. \(2^{58}\)
• C. \(2^{30}\)
• D. \(2^{29}\)

Hint:

For odd \(n\), binomial coefficients are symmetric, and the sum of the first half equals the sum of the second half.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The coefficients are binomial coefficients \(\binom{59}{r}\) for \(r = 0\) to \(59\). The sum of all coefficients is \(2^{59}\). The last 30 coefficients correspond to \(r = 30\) to \(59\).

Step 2: Detailed Explanation:
Total coefficients = \(2^{59}\). Since \(\binom{59}{r} = \binom{59}{59-r}\), the first 30 coefficients (r=0 to 29) are equal to the last 30 coefficients (r=30 to 59). The middle coefficient \(\binom{59}{29.5}\) is not integer since 59 is odd; the two middle terms are \(\binom{59}{29}\) and \(\binom{59}{30}\), which are equal. Let \(S\) = sum of last 30 coefficients = sum of first 30 coefficients. Then \(2S = 2^{59} - \binom{59}{29.5}\)? Since total is \(2^{59}\), and the middle term is not counted twice. For odd \(n=59\), there are 60 terms, symmetric. The sum of first 30 terms equals sum of last 30 terms. So total = \(2S = 2^{59}\). Thus \(S = 2^{58}\).

Step 3: Final Answer:
Sum = \(2^{58}\), which corresponds to option (B).