Problem 14 - Entrance Test

The harmonic mean of n numbers x_1, x_2, ..., x_n is H = n / (Σ 1/x_i). For the first n natural numbers, H_n = n / (1 + 1/2 + 1/3 + ... + 1/n) = n / H_n', where H_n' is the nth harmonic number. The sum of reciprocals is the nth harmonic number: h_n = Σ_{i=1}^n 1/i. So H_n = n / h_n. The problem states H_n = k/(n+1). So n / h_n = k/(n+1) → k = n(n+1)/h_n. But h_n does not simplify to a nice expression in general. However, for specific n, k is an integer. Let me compute for small n. n=1: H_1 = 1, k/(2) = 1 → k=2. n=2: numbers 1,2. HM = 2/(1+1/2) = 2/(3/2) = 4/3. k/3 = 4/3 → k=4. n=3: HM = 3/(1+1/2+1/3) = 3/(11/6) = 18/11. k/4 = 18/11 → k=72/11, not integer. So k is not always integer. Perhaps the problem means something else. Maybe H_n denotes the nth harmonic number (sum of reciprocals), not the harmonic mean. If H_n is the nth harmonic number, then H_n = 1 + 1/2 + ... + 1/n. And H_n = k/(n+1). This is not true in general. Let me reconsider: Perhaps the problem means the harmonic mean of the first n natural numbers can be expressed as k/(n+1). For n=1: HM=1=2/2, k=2. n=2: HM=4/3=4/3, k=4. n=3: HM=18/11, which is not of the form k/4. So only for n=1,2. This doesn't make sense. Perhaps the problem is: The harmonic mean of the first n natural numbers is equal to n(n+1)/(2H_n) where H_n is the nth harmonic number? No. Let me re-interpret: 'The harmonic mean of the first n natural numbers is H_n. If H_n = k / (n+1) for some integer k, then k is:' Perhaps H_n here denotes the nth harmonic number (the sum), and the harmonic mean is something else. The harmonic mean of 1,2,...,n is n / H_n. If this equals k/(n+1), then n/H_n = k/(n+1) → k = n(n+1)/H_n. For n=1: H_1=1, k=1*2/1=2. For n=2: H_2=1+1/2=3/2, k=2*3/(3/2)=6/(3/2)=4. For n=3: H_3=11/6, k=3*4/(11/6)=12*6/11=72/11. Not integer. So k is not always integer. Given the options, the intended answer is likely n(n+1) (Option B) or 2n(n+1) (Option D). Let me select Option B.