Question

Show that for all positive integer values of n 2^n+1+2^n+2+2^n+3 is divisible by 7

Answer 1

`2^(n+1) + 2^(n+2) + 2^(n+3) = 2^(n + 1) * 7`

Since there is a factor of 7, it is divisible by 7.

Explanation:

`2^(n+1) + 2^(n+2) + 2^(n+3) =`

`= 2^(n + 1) * 2^0 * 2^(n + 1) * 2^1 + 2^(n + 1) * 2^2`

`= 2^(n + 1) * 1 * 2^(n + 1) * 2 + 2^(n + 1) * 3`

`= 2^(n + 1) * (1 + 2 + 3)`

`= 2^(n + 1) * 7`

Since there is a factor of 7, it is divisible by 7.

Answer 2

the expression 2^n+1 + 2^n+2 + 2^n+3 is divisible by 7.

Explanation:

Step 1: Base case

Let's check the expression for the smallest possible value of n, which is n = 1:

2^1+1 + 2^1+2 + 2^1+3 = 2^2 + 2^3 + 2^4 = 4 + 8 + 16 = 28.

Since 28 is divisible by 7, the base case holds.

Step 2: Inductive hypothesis

Assume that for some positive integer k, the expression 2^k+1 + 2^k+2 + 2^k+3 is divisible by 7.

Step 3: Inductive step

We need to prove that if the hypothesis holds for k, it also holds for k+1.

For k+1:

2^(k+1)+1 + 2^(k+1)+2 + 2^(k+1)+3

= 2^(k+1) * 2^1 + 2^(k+1) * 2^2 + 2^(k+1) * 2^3

= 2^k * 2 + 2^k * 4 + 2^k * 8

= 2^k * (2 + 4 + 8)

= 2^k * 14.

Now, we can rewrite 2^k * 14 as 2^k * (2 * 7). Since 2^k is an integer and 7 is a prime number, we know that 2^k * 7 is divisible by 7.

Therefore, we have shown that if the hypothesis holds for k, it also holds for k+1.

Step 4: Conclusion

By the principle of mathematical induction, we have proven that for all positive integer values of n, the expression 2^n+1 + 2^n+2 + 2^n+3 is divisible by 7.