If `a_(n) = 2^(2^(n)) + 1`, then for n > 1, last digit of `a_(n)` is

Options

A. 5
B. 7
C. 3
D. 4

Answer

To find the last digit of \( a_n = 2^{2^n} + 1 \) for \( n > 1 \), we will analyze the expression step by step.

### Step 1: Calculate \( a_2 \)
Let's start by calculating \( a_2 \):
\[ a_2 = 2^{2^2} + 1 = 2^4 + 1 = 16 + 1 = 17 \]
The last digit of \( 17 \) is \( 7 \).

### Step 2: Assume it holds for \( n = k \)
Now, let's assume that for some integer \( k > 1 \), the last digit of \( a_k = 2^{2^k} + 1 \) is \( 7 \). This means:
\[ a_k \equiv 7 \mod 10 \]
or equivalently,
\[ a_k = 10m + 7 \text{ for some integer } m. \]

### Step 3: Prove it for \( n = k + 1 \)
Now we need to prove that \( a_{k+1} \) also has a last digit of \( 7 \):
\[ a_{k+1} = 2^{2^{k+1}} + 1 = 2^{2 \cdot 2^k} + 1 = (2^{2^k})^2 + 1. \]
Let \( x = 2^{2^k} \). Then we have:
\[ a_{k+1} = x^2 + 1. \]

### Step 4: Find the last digit of \( x^2 + 1 \)
Since we assumed \( a_k \equiv 7 \mod 10 \), we know:
\[ x \equiv 2^{2^k} \mod 10. \]
We need to find the last digit of \( x \). The last digits of powers of \( 2 \) cycle every \( 4 \):
- \( 2^1 \equiv 2 \)
- \( 2^2 \equiv 4 \)
- \( 2^3 \equiv 8 \)
- \( 2^4 \equiv 6 \)
- \( 2^5 \equiv 2 \) (and so on)

To find \( 2^{2^k} \mod 10 \), we need \( 2^k \mod 4 \):
- If \( k \equiv 0 \mod 4 \), then \( 2^{2^k} \equiv 6 \mod 10 \).
- If \( k \equiv 1 \mod 4 \), then \( 2^{2^k} \equiv 2 \mod 10 \).
- If \( k \equiv 2 \mod 4 \), then \( 2^{2^k} \equiv 4 \mod 10 \).
- If \( k \equiv 3 \mod 4 \), then \( 2^{2^k} \equiv 8 \mod 10 \).

### Step 5: Calculate \( x^2 + 1 \)
Now we calculate \( x^2 + 1 \) for each case:
- If \( x \equiv 2 \mod 10 \), then \( x^2 + 1 \equiv 4 + 1 \equiv 5 \mod 10 \).
- If \( x \equiv 4 \mod 10 \), then \( x^2 + 1 \equiv 16 + 1 \equiv 7 \mod 10 \).
- If \( x \equiv 6 \mod 10 \), then \( x^2 + 1 \equiv 36 + 1 \equiv 7 \mod 10 \).
- If \( x \equiv 8 \mod 10 \), then \( x^2 + 1 \equiv 64 + 1 \equiv 5 \mod 10 \).

### Conclusion
Thus, for \( n > 1 \), the last digit of \( a_n \) is consistently \( 7 \) when \( n \equiv 2 \) or \( n \equiv 3 \mod 4 \).

Therefore, the last digit of \( a_n \) for \( n > 1 \) is:
\[ \boxed{7} \]