Questions: Which explicit formula is equivalent to a1=1, an=4 an-1 ? an=4(4)^n-1 an=4(1)^n-1 an=1(4)^n-1 an=1+(n-1) 4

Solution Steps

The given recurrence relation is: \[ a_1 = 1, \quad a_n = 4a_{n-1} \]

This means the first term \( a_1 \) is 1, and each subsequent term is 4 times the previous term.

To find the explicit formula, we need to express \( a_n \) in terms of \( n \) without using the previous term \( a_{n-1} \).

Starting with the given recurrence relation:

• \( a_1 = 1 \)
• \( a_2 = 4a_1 = 4 \times 1 = 4 \)
• \( a_3 = 4a_2 = 4 \times 4 = 16 \)
• \( a_4 = 4a_3 = 4 \times 16 = 64 \)

We observe that: \[ a_n = 4^{n-1} \times a_1 = 4^{n-1} \times 1 = 4^{n-1} \]

Thus, the explicit formula is: \[ a_n = 1 \times 4^{n-1} \]

Now, we compare the derived explicit formula \( a_n = 1 \times 4^{n-1} \) with the given options:

• \( a_n = 4(4)^{n-1} \)
• \( a_n = 4(1)^{n-1} \)
• \( a_n = 1(4)^{n-1} \)
• \( a_n = 1+(n-1)4 \)

The correct match is: \[ a_n = 1(4)^{n-1} \]

Final Answer