Question

If A = {62n -35n -1: n = 1,2,3,...} and B = {35(n-1) : n = 1,2,3,...} then which of the following is true?

• A. Neither every member of A is in B nor every member of B is in A
• B. Every member of A is in B and at least one member of B is not in A
• C. Every member of B is in A.
• D. At least one member of A is not in B

Answer 1

The Correct Option is B

Step 1: Define the Expression A 

We are given:

\[ A = 36^n - 35^n - 1 = (36^n - 1^n) - 35^n \]

Let’s write it as:

\[ A = 36^n - 1^n - 35^n \]

Step 2: Use Algebraic Property of Exponents

We use the identity:

\[ a^n - b^n \text{ is divisible by } (a - b) \text{ for all positive integers } n \]

Therefore,

\[ 36^n - 1^n \text{ is divisible by } 36 - 1 = 35 \]

So, the expression \( A = (36^n - 1^n) - 35^n \) is divisible by 35 minus \( 35^n \). But since:

\[ 36^n - 1^n \equiv 0 \pmod{35} \Rightarrow A = 36^n - 1^n - 35^n \equiv -35^n \pmod{35} \]

Now note that \( 35^n \equiv 0 \pmod{35} \Rightarrow A \equiv 0 \pmod{35} \)

Thus, for any integer \( n \), the expression \( A \) is divisible by 35.

Step 3: Define the Sets

Let:

• Set \( A \) contain all values of the expression \( 36^n - 1^n - 35^n \) for positive integers \( n \)
• Set \( B \) = all multiples of 35 including 0

Step 4: Conclusion on Set Relationship

Every value of the expression \( A \) is divisible by 35, hence:

\[ A \subseteq B \]

But not all multiples of 35 (e.g., 35, 70, 105, ...) can be expressed as \( 36^n - 1^n - 35^n \), so:

\[ B \not\subseteq A \Rightarrow A \subset B \]

Final Answer:

\[ \boxed{\text{Every member of A is in B, but not every member of B is in A}} \]