Question

Prove the following statements with either induction, strong induction or proof by smallest counterexample. Concerning the Fibonacci sequence, prove that \(F_{2}+F_{4}+F_{6}+F_{8}+\cdots+F_{2 n}=F_{2 n+1}-1 .\)

Short answer

The proposition, \(F_{2}+F_{4}+F_{6}+...+F_{2n} = F_{2n+1}-1\), is proven to be true for all positive integers \(n\) by using the method of mathematical induction.

Step-by-step solution

Prove Base Case

The base case is when \(n = 1\). In this case, the left side of the equation is \(F_2 = 1\), while the right side is \(F_3 - 1 = 1\). Therefore, the equation holds true for \(n = 1\).

Assume the Statement is True for Some \(k\)

Assume that the formula is true for \(n = k\), which is \(F_2 + F_4 + \cdots + F_{2k} = F_{2k+1} - 1\). This is our induction hypothesis.

Prove the Statement is True for \(n=k+1\)

If our formula is true for \(n=k\), then we need to prove it's true for \(n=k+1\). Let's add \(F_{2k+2}\) on both sides of our induction hypothesis. We get \(F_2 + F_4 + \cdots + F_{2k} + F_{2k+2} = F_{2k+1} -1 + F_{2k+2}\). On the right side of this equation, using Fibonacci property \(F_{n} = F_{n-1} + F_{n-2}\), we simplify it to \(F_{2k+3} - 1\). So, \(F_2 + F_4 + \cdots + F_{2k} + F_{2k+2} = F_{2(k+1) +1} -1\). This shows the formula is true for \(n=k+1\) if it's true for \(n=k\).

Conclusion

Thus, because the formula holds for \(n=1\) (the Base case), and if it is true for \(n=k\) then it is also true for \(n=k+1\). We conclude that the proposition, \(F_{2}+F_{4}+F_{6}+\cdots+F_{2n} = F_{2n+1}-1\), is true for all positive integers \(n\).