Find the sum of the last 30 coefficients in the expansion of `(1+x)^(59),` when expanded in ascending powers of `x`.

Explanation

To find the sum of the last 30 coefficients in the expansion of \((1+x)^{59}\), we can follow these steps:

### Step 1: Understand the Binomial Expansion
The binomial expansion of \((1+x)^n\) is given by:
\[ (1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]
For our case, \(n = 59\), so:
\[ (1+x)^{59} = \sum_{k=0}^{59} \binom{59}{k} x^k \]

### Step 2: Identify the Last 30 Coefficients
The last 30 coefficients in the expansion correspond to the terms from \(k=30\) to \(k=59\). Therefore, we need to find:
\[ S = \binom{59}{30} + \binom{59}{31} + \binom{59}{32} + \ldots + \binom{59}{59} \]

### Step 3: Use the Symmetry of Binomial Coefficients
We can use the property of binomial coefficients:
\[ \binom{n}{k} = \binom{n}{n-k} \]
Thus, we can rewrite \(S\) as:
\[ S = \binom{59}{30} + \binom{59}{31} + \ldots + \binom{59}{59} = \binom{59}{29} + \binom{59}{28} + \ldots + \binom{59}{0} \]

### Step 4: Add the Two Expressions
Now, we can add the two expressions for \(S\):
\[ 2S = \left( \binom{59}{0} + \binom{59}{1} + \ldots + \binom{59}{29} \right) + \left( \binom{59}{30} + \binom{59}{31} + \ldots + \binom{59}{59} \right) \]
This simplifies to:
\[ 2S = \binom{59}{0} + \binom{59}{1} + \ldots + \binom{59}{59} \]

### Step 5: Calculate the Total Sum of Coefficients
The sum of all coefficients in the binomial expansion \((1+x)^{59}\) is given by:
\[ 2^{59} \]
Thus:
\[ 2S = 2^{59} \]

### Step 6: Solve for \(S\)
Now, we can solve for \(S\):
\[ S = \frac{2^{59}}{2} = 2^{58} \]

### Final Answer
The sum of the last 30 coefficients in the expansion of \((1+x)^{59}\) is:
\[ \boxed{2^{58}} \]