Let `t_r=2^(r/2)+2^(-r/2)` then `sum_(r=1)^10 t_r^2=`

Options

A. `(2^(21)-1)/(2^(10))+20`
B. `(2^(21)-1)/(2^(10))+19`
C. `(2^(21)-1)/20-1`
D. none of these

Answer

To solve the problem, we need to find the sum of \( t_r^2 \) from \( r = 1 \) to \( r = 10 \), where \( t_r = 2^{r/2} + 2^{-r/2} \).

### Step-by-Step Solution:

1. **Define \( t_r \)**:
\[ t_r = 2^{r/2} + 2^{-r/2} \]

2. **Calculate \( t_r^2 \)**:
\[ t_r^2 = (2^{r/2} + 2^{-r/2})^2 \]
Expanding this using the formula \( (a + b)^2 = a^2 + 2ab + b^2 \):
\[ t_r^2 = (2^{r/2})^2 + 2 \cdot (2^{r/2})(2^{-r/2}) + (2^{-r/2})^2 \]
Simplifying each term:
\[ t_r^2 = 2^r + 2 + 2^{-r} \]

3. **Set up the summation**:
We need to find:
\[ \sum_{r=1}^{10} t_r^2 = \sum_{r=1}^{10} (2^r + 2 + 2^{-r}) \]
This can be separated into three sums:
\[ \sum_{r=1}^{10} t_r^2 = \sum_{r=1}^{10} 2^r + \sum_{r=1}^{10} 2 + \sum_{r=1}^{10} 2^{-r} \]

4. **Calculate each sum**:
- **First Sum**: \( \sum_{r=1}^{10} 2^r \) is a geometric series:
\[ \sum_{r=1}^{10} 2^r = 2 \cdot \frac{2^{10} - 1}{2 - 1} = 2(2^{10} - 1) = 2^{11} - 2 \]

- **Second Sum**: \( \sum_{r=1}^{10} 2 = 2 \cdot 10 = 20 \)

- **Third Sum**: \( \sum_{r=1}^{10} 2^{-r} \) is also a geometric series:
\[ \sum_{r=1}^{10} 2^{-r} = \frac{1/2}{1 - 1/2} \left(1 - (1/2)^{10}\right) = 1 - \frac{1}{2^{10}} = 1 - \frac{1}{1024} = \frac{1023}{1024} \]

5. **Combine the results**:
Now, we can combine all the sums:
\[ \sum_{r=1}^{10} t_r^2 = (2^{11} - 2) + 20 + \frac{1023}{1024} \]
Simplifying this:
\[ = 2^{11} + 18 + \frac{1023}{1024} \]
Converting \( 18 \) to a fraction:
\[ = 2^{11} + \frac{18432}{1024} + \frac{1023}{1024} = 2^{11} + \frac{18432 + 1023}{1024} = 2^{11} + \frac{19455}{1024} \]

### Final Answer:
Thus, the final result is:
\[ \sum_{r=1}^{10} t_r^2 = 2^{11} + \frac{19455}{1024} \]