Question

9^n-8^n-1 is divisible by 8 ​

Answer 1

you can't get any integer but this can be written

Answer 2

Concept:

According to the Principle of Induction, a mathematical statement P(n) is true for all natural number n if it satisfies below two conditions.

• P(n) is true for n = 1.
• P(n) is true for n=k+1, provided P(k) is true.

Given:

The given statement is "$9^n-8^n-1$ is divisible by 8".

Find:

We have to prove that the above statement is true for all natural number n.  

Solution:

Let the statement P(n) be "$9^n-8^n-1$ is divisible by 8".

When n=1, $P(1)=9^1-8^1-1=9-8-1=0$

P(n) is true for n=1 since 0 is divisible by 8.

When n=2, $P(2)=9^2-8^2-1=81-64-1=16$

P(n) is true for n=2 since 16 is divisible by 8.

Let P(k) is true.

So, $9^k-8^k-1=8M$, where M is constant.

Now for n=k+1, we get,

$P(k+1)=9^{k+1}-8^{k+1}-1\\~~~~~~~~~~~~~=9.9^k-8.8^k-1\\~~~~~~~~~~~~~=9(9^k-8^k-1)+9.8^k+9-8.8^k-1\\~~~~~~~~~~~~~=9\cdot8M+(9-8).8^k+8\\~~~~~~~~~~~~~=9\cdot8M+8^k+8\\~~~~~~~~~~~~~=8(9M+8^{k-1}+1)\\~~~~~~~~~~~~~=8L$

where L is also a constant

So clearly P(k+1) is also true.

By the principle of induction P(n) is true for all natural number n.

Hence the statement "$\mathbf{9^n-8^n-1}$ is divisible by 8" is true for all natural number n.

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