Question

Use mathematical induction to prove that \(9\) divides \({n^3} + {\left( {n + 1} \right)^3} + {\left( {n + 2} \right)^3}\) whenever \(n\)is a nonnegative integer.

Short answer

It is proved that\(9\)divides\({n^3} + {\left( {n + 1} \right)^3} + {\left( {n + 2} \right)^3}\)whenever nis a nonnegative integer.

Step-by-step solution

Principle of Mathematical Induction

Consider the propositional function\(P\left( n \right)\). Consider two actions to prove that\(P\left( n \right)\)evaluates to accurate for all set of positive integers\(n\).

Consider the first basic step is to confirm that \(P\left( 1 \right)\)true.

Consider the inductive step is to demonstrate that for any positive integer k the conditional statement \[P\left( k \right) \to P\left( {k + 1} \right)\]is true.

 Step 2: Prove the basis step

Let,\(P\left( n \right)\): “\(9\) divides\({n^3} + {\left( {n + 1} \right)^3} + {\left( {n + 2} \right)^3}\)”

In the basis step, we need to prove that\(P\left( 1 \right)\)is true.

For finding statement\(P\left( 1 \right)\)substituting\(1\)for\(n\)in the statement.

\(\begin{array}{c}{n^3} + {\left( {n + 1} \right)^3} + {\left( {n + 2} \right)^3} = {1^3} + {\left( {1 + 1} \right)^3} + {\left( {1 + 2} \right)^3}\\ = 1 + 8 + 27\\ = 36\end{array}\)

It is known that\(36\)is divisible by 9.

From the above, we can see that the statement \(P\left( 1 \right)\) is true this is also known as the basis step of the proof.

Prove the Inductive step

In the inductive step, we need to prove that, if\(P\left( k \right)\)is true, then\(P\left( {k + 1} \right)\)is also true.

That is,

\(P\left( k \right) \to P\left( {k + 1} \right)\)is true for all positive integers k.

In the inductive hypothesis, we assume that\(P\left( k \right)\)is true for any arbitrary positive integer\(k\)

That is\(9\)divides\({k^3} + {\left( {k + 1} \right)^3} + {\left( {k + 2} \right)^3}\).

Now show that\(P\left( {k + 1} \right)\)is also true.

Therefore, replace\(k\)with\(k + 1\)in the statement.

\(\begin{array}{c}{\left( {k + 1} \right)^3} + {\left( {k + 1 + 1} \right)^3} + {\left( {k + 1 + 2} \right)^3} = {\left( {k + 1} \right)^3} + {\left( {k + 2} \right)^3} + {\left( {k + 3} \right)^3}\\ = {\left( {k + 1} \right)^3} + {\left( {k + 2} \right)^3} + {k^3} + 9{k^2} + 9k + 27\\ = \left[ {{{\left( {k + 1} \right)}^3} + {{\left( {k + 2} \right)}^3} + {k^3}} \right] + 9\left[ {{k^2} + k + 3} \right]\end{array}\)

The first term of the expression is divisible by 9 from the inductive hypothesis and second term is also divisible by 9.

Since both the terms are divisible by 9, therefore\({\left( {k + 1} \right)^3} + {\left( {k + 1 + 1} \right)^3} + {\left( {k + 1 + 2} \right)^3}\)

Is also divisible by 9.

From the above, it is seen that\(P\left( {k + 1} \right)\)is also true.

Hence,\(P\left( {k + 1} \right)\)is true under the assumption that\(P\left( k \right)\)is true. This

completes the inductive step.

Hence, it is proved that\(9\)divides\({n^3} + {\left( {n + 1} \right)^3} + {\left( {n + 2} \right)^3}\)whenever nis a nonnegative integer.