Question

I have a question
I need absolutely correct answer​

Answer 1

Answer: METHOD 1

Firstly, for a number to be divisible 9, obviously it must also be divisible by 3

So it is sufficient to prove that it divisible by 9 and this will automatically imply that it is divisible by 3

Proof: By induction

We quickly plug in few values of n

n=1 , and we have:

3•4¹ +51=63 =9×7 ( divisible by 9)

n=2

3•4²+51=99=9×11 ( again divisible by 9)

By Inductive Hypothesis:

We assume it holds for n=k

Thus:

3•4^k +51 =9m ( where m = 1,2,3…..)

Hence we prove it holds for n=k+1

Thus we have:

3•4^(k+1) +51 => 3•4•4^k +51

=> ( 3•4^k +51) + (3•3•4^k)

=> 9(m + 4^k)

This completes the induction and hence completes the proof

Step-by-step explanation: METHOD 2

3 * 4^n + 51 = 3 * (4^n + 17),

which is divisible by 3.

I will prove that 4^n + 17 is divisible by 3

Case 1: n = 1

4^1 + 17 = 21 , which is divisible by 3.

Let n = k, a positive integer, such that

4^k + 17 is divisible by 3

If n = k + 1, then

4^(k + 1) + 17 = 4 * 4^k + 17

= (4^k + 17) + 3 * 4^k

4^k + 17 and 3 * 4^k are both divisible by 3,

so 4^k + 17 + 3 * 4^k is divisible by 3 as well.

Since 4^n + 17 is divisible by 3 for any positive integer n, then

3 * (4^n + 17) would be divisible by 9.