Question

Show that \(l_0^2 + l_1^2 + ... + l_n^2 = {l_n}{l_{n + 1}} + 2\)whenever nis a nonnegative integer and \({l_i}\)is the \(ith\) Lucas number.

Short answer

It is proven that\(l_0^2 + l_1^2 + ... + l_n^2 = {l_n}{l_{n + 1}} + 2\).

Step-by-step solution

Principle of Mathematical Induction

To prove that\(P\left( n \right)\)is true for all positive integers n, where\(P\left( n \right)\)is a propositional function, it completes two steps:

Basis Step:

It verifies that \(P\left( 1 \right)\)is true.

Inductive Step:

It shows that the conditional statement \(P\left( k \right) \to P\left( {k + 1} \right)\) is true for all positive integers k.

Lucas numbers

Lucas numbers are defined by \({l_0} = 2,\,\,{l_1} = 1,\)and \({l_n} = {l_{n - 1}} + {l_{n - 2}}\) for \(n = 2,\,3,\,4,....\).

Proving the basis step

Let\(P\left( n \right)\): “\(l_0^2 + l_1^2 + ... + l_n^2 = {l_n}{l_{n + 1}} + 2\)”

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

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

\(\begin{aligned}{c}l_0^2 + l_1^2 &= {l_1}{l_{1 + 1}} + 2\\{2^2} + {1^2} &= {l_1}\left( {{l_1} + {l_0}} \right) + 2\\5 &= 1\left( {1 + 2} \right) + 2\\5 &= 5\end{aligned}\)

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

Proving the Inductive step

In the inductive step, it needs 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, it assumes that\(P\left( k \right)\)is true for any arbitrary positive integer\(k\).

That is,\(l_0^2 + l_1^2 + ... + l_k^2 = {l_k}{l_{k + 1}} + 2\).

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

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

\(l_0^2 + l_1^2 + ... + l_{k + 1}^2 = l_0^2 + l_1^2 + ... + l_k^2 + l_{k + 1}^2\)

From the inductive hypothesis,\(l_0^2 + l_1^2 + ... + l_k^2 = {l_k}{l_{k + 1}} + 2\).

Therefore, it is written as:

\(\begin{aligned}{c}l_0^2 + l_1^2 + ... + l_{k + 1}^2 &= {l_k}{l_{k + 1}} + 2 + l_{k + 1}^2\\ &= {l_k}{l_{k + 1}} + l_{k + 1}^2 + 2\\ &= {l_{k + 1}}\left( {{l_k} + {l_{k + 1}}} \right) + 2\\ &= {l_{k + 1}}{l_{k + 2}} + 2\end{aligned}\)

From the above, it can see 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 proven that\(l_0^2 + l_1^2 + ... + l_n^2 = {l_n}{l_{n + 1}} + 2\).