Question

PLEASE ANSWER THIS QUESTION......FROM MATHEMATICAL INDUCTION​

Answer 1

Answer:

Step-by-step explanation:

Let P(n) be the statement that " 5ⁿ + 2×3ⁿ⁻¹ + 1 is divisible by 8 ".

There are two steps:

1. Show that P(1) is true.
2. Ahow that if P(k) is true, then P(k+1) is true.

Step 1

Putting n=1, the statement P(1) is " 5¹ + 2×3⁰ + 1 is divisible by 8 ".

5¹ + 2×3⁰ + 1 = 5 + 2×1 + 1 = 5 + 2 + 1 = 8, which is certainly divisible by 8.

So P(1) is true.

Step 2

Suppose for some k ≥1, the statement P(k) is true: " 5^k + 2×3^(k-1) + 1 is divisible by 8 ".

We must use this to establish that P(k+1) must then be true, too.

The statement P(k+1) is " 5^(k+1) + 2×3^k + 1  is divisible by 8 ".

5^(k+1) + 2×3^k + 1

= 5×5^k + 6×3^(k-1) + 1

= 5×5^k + 10×3^(k-1) + 5 - 4×3^(k-1) - 4

= 5×(5^k + 2×3^(k-1) + 1) - 4×( 3^(k-1) + 1 )           ... (*)

Since P(k) is true (the inductive hypothesis), the first term in (*) is divisible by 8.  Since k ≥ 1, the value 3^(k-1) is odd  =>  3^(k-1) + 1 is even  => the second term in (*) is divisible by 8.

Since both terms in (*) are divisible by 8, the whole expression is divisible by 8.  It follows that 5^(k+1) + 2×3^k + 1 is divisible by 8, and so P(k+1) is true.

The claim now follows by mathematical induction.

Answer 2

let's us consider

s(n)=