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

To find the sum of the last 30 coefficients in the expansion of \((1 + x)^{59}\), we need to consider some properties of binomial expansions.

The binomial theorem states that:

\((1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k\)

Here, \(n = 59\). Therefore, the expansion will be:

\((1 + x)^{59} = \binom{59}{0} + \binom{59}{1}x + \binom{59}{2}x^2 + \ldots + \binom{59}{59}x^{59}\)

We are interested in the sum of the last 30 coefficients, i.e., coefficients from \(\binom{59}{30}\) to \(\binom{59}{59}\). According to the symmetry property of binomial coefficients:

\(\binom{n}{k} = \binom{n}{n-k}\)

This allows us to write:

• \(\binom{59}{30} = \binom{59}{29}\)
• \(\binom{59}{31} = \binom{59}{28}\)
• \(\ldots\)
• \(\binom{59}{59} = \binom{59}{0}\)

As such, the sum of the last 30 coefficients is equivalent to the sum of the first 30 coefficients. Thus, we obtain the following equality:

\(\sum_{k=30}^{59} \binom{59}{k} = \sum_{k=0}^{29} \binom{59}{k}\)

The sum of all the coefficients in any binomial expansion \((1+x)^n\) is:

\((1+1)^n = 2^n\)

Hence, for our case with \(n = 59\):

\(\sum_{k=0}^{59} \binom{59}{k} = 2^{59}\)

We know that:

\(\sum_{k=0}^{29} \binom{59}{k} + \sum_{k=30}^{59} \binom{59}{k} = 2^{59}\)

Since both these sums are equal, each is half of \(2^{59}\):

\(\sum_{k=0}^{29} \binom{59}{k} = \sum_{k=30}^{59} \binom{59}{k} = \frac{2^{59}}{2} = 2^{58}\)

Thus, the sum of the last 30 coefficients is \(2^{58}\).

Therefore, the correct answer is:

\(2^{58}\)