PutnamSuperchallengeIntegrals I: Feynman's Trick ShareIntegrals I: Feynman's Trick · Superchallenge problemDefine f(1)=1f(1) = 1f(1)=1 and f(n)=f(⌊n/2⌋)+1f(n) = f(\lfloor n/2\rfloor) + 1f(n)=f(⌊n/2⌋)+1 for n≥2n \ge 2n≥2. Strong induction proves f(n)=⌊log2n⌋+1f(n) = \lfloor\log_2 n\rfloor + 1f(n)=⌊log2n⌋+1. Compute ∑n=1100f(n)\displaystyle\sum_{n=1}^{100} f(n)n=1∑100f(n).Check Show answer Walk through it step by step Practice more like this Review this concept