Thread: Proving recursive function via mathematical induction (EASY, but need a boost)

This was a sample problem discussed in class that I already have the solution to, but I am having trouble understanding the highlighted section. Please be very detailed with your explanation. Thank you in advance!

Let f be defined recursively by:
Initial condition: f(0)=1
Recursive step: f(n)=2f(n-1)+1

For the function f above, prove that for all n, f(n)=2ⁿ⁺¹-1.

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SOLUTION:

Proof by induction:
1) Basis step:
Prove that f(0)=2ⁿ⁺¹-1
f(0)=2⁰⁺¹-1 is true
because 1=1
2) Inductive step:
Inductive hypothesis: Assume that for some n>=0, f(n)=2ⁿ⁺¹-1.
We need to use this hypothesis to prove that f(n+1)=2ⁿ⁺²-1.
Proof: Assume that the inductive hypothesis is true,
(I DO NOT UNDERSTAND HOW THE RIGHT SIDE OF THE BELOW EQUATION (2f(n)+1) WAS FOUND)
then f(n+1)=2f(n)+1 (via Recursive definition of f)
=2(2ⁿ⁺¹-1)+1 (via Inductive hypothesis)
=2ⁿ⁺²-2+1
=2ⁿ⁺¹-1

# END OF PROOF

After having shown the base case is true, I would state the induction hypothesis :



Using the recursive definition, we may state:









We have derived from , thereby completing the proof by induction.

You compose the functions on each side of the equation of the recursive step with g(k)=k+1 and obtain the second equation in red with k instead of n.

Sorry for the late reply. Thank you!