Find the sum of last 20 coefficients in the expansions of `(1+x)^(39).`

Answer

To find the sum of the last 20 coefficients in the expansion of \((1+x)^{39}\), we can follow these steps:

### Step 1: Understand the Binomial Expansion
The binomial expansion of \((1+x)^{39}\) is given by:
\[ (1+x)^{39} = \sum_{k=0}^{39} \binom{39}{k} x^k \]
where \(\binom{39}{k}\) are the binomial coefficients.

### Step 2: Identify the Last 20 Coefficients
The last 20 coefficients in this expansion correspond to the terms from \(k=20\) to \(k=39\). Therefore, we need to calculate:
\[ \sum_{k=20}^{39} \binom{39}{k} \]

### Step 3: Use the Property of Binomial Coefficients
We can use the property of binomial coefficients that states:
\[ \sum_{k=0}^{n} \binom{n}{k} = 2^n \]
For \(n=39\), we have:
\[ \sum_{k=0}^{39} \binom{39}{k} = 2^{39} \]

### Step 4: Relate the Coefficients
We can also express the sum of the last 20 coefficients as:
\[ \sum_{k=20}^{39} \binom{39}{k} = \sum_{k=0}^{39} \binom{39}{k} - \sum_{k=0}^{19} \binom{39}{k} \]
Thus, we can write:
\[ \sum_{k=20}^{39} \binom{39}{k} = 2^{39} - \sum_{k=0}^{19} \binom{39}{k} \]

### Step 5: Use the Symmetry of Binomial Coefficients
Using the symmetry property \(\binom{n}{k} = \binom{n}{n-k}\), we can relate \(\sum_{k=0}^{19} \binom{39}{k}\) to \(\sum_{k=20}^{39} \binom{39}{k}\):
\[ \sum_{k=0}^{19} \binom{39}{k} = \sum_{k=20}^{39} \binom{39}{k} \]
This means:
\[ \sum_{k=20}^{39} \binom{39}{k} = \sum_{k=0}^{19} \binom{39}{k} \]

### Step 6: Set Up the Equation
Let \(S = \sum_{k=20}^{39} \binom{39}{k}\). Then we have:
\[ S = 2^{39} - S \]
This simplifies to:
\[ 2S = 2^{39} \]
Thus:
\[ S = 2^{38} \]

### Conclusion
The sum of the last 20 coefficients in the expansion of \((1+x)^{39}\) is:
\[ \boxed{2^{38}} \]