`A_(n)=2^(n)-1` for all integers `n ge1`
`{:("Quantity A","Quantity B"),("The units digit of "A_(26),"The units digit of "A_(34)):}`

Answer

To solve the problem, we need to find the units digits of \( A_{26} \) and \( A_{34} \), where \( A_n = 2^n - 1 \).

### Step-by-Step Solution:

1. **Identify the formula**:
We have \( A_n = 2^n - 1 \).

2. **Find the units digit of \( A_{26} \)**:
- First, calculate \( 2^{26} \).
- The units digits of powers of 2 follow a repeating pattern:
- \( 2^1 = 2 \) (units digit 2)
- \( 2^2 = 4 \) (units digit 4)
- \( 2^3 = 8 \) (units digit 8)
- \( 2^4 = 16 \) (units digit 6)
- Then it repeats: 2, 4, 8, 6...
- The pattern repeats every 4 terms. To find the units digit of \( 2^{26} \), we calculate \( 26 \mod 4 \):
\[ 26 \div 4 = 6 \quad \text{(remainder 2)} \]
- Since the remainder is 2, the units digit of \( 2^{26} \) is the same as that of \( 2^2 \), which is 4.
- Therefore, the units digit of \( A_{26} = 2^{26} - 1 \) is:
\[ 4 - 1 = 3 \]

3. **Find the units digit of \( A_{34} \)**:
- Similarly, calculate \( 2^{34} \).
- Again, using the same pattern, we find \( 34 \mod 4 \):
\[ 34 \div 4 = 8 \quad \text{(remainder 2)} \]
- The remainder is also 2, so the units digit of \( 2^{34} \) is the same as that of \( 2^2 \), which is 4.
- Therefore, the units digit of \( A_{34} = 2^{34} - 1 \) is:
\[ 4 - 1 = 3 \]

4. **Compare the results**:
- The units digit of \( A_{26} \) is 3.
- The units digit of \( A_{34} \) is also 3.

### Conclusion:
Since both quantities have the same units digit, we conclude:
\[ \text{Quantity A} = \text{Quantity B} \]