{
 "aufgabe": "Compute\n\\[\\sum_{n = 1}^\\infty \\frac{F_{n + 1}}{F_n F_{n + 2}},\\]where $F_n$ denotes the $n$th Fibonacci number, so $F_0 = 0$ and $F_1 = 1.$",
 "musterloesung": "Since $F_{n + 1} = F_{n + 2} - F_n,$\n\\[\\frac{F_{n + 1}}{F_n F_{n + 2}} = \\frac{F_{n + 2} - F_n}{F_n F_{n + 2}} = \\frac{1}{F_n} - \\frac{1}{F_{n + 2}}.\\]Then\n\\begin{align*}\n\\sum_{n = 1}^\\infty \\frac{1}{F_n F_{n + 2}} &= \\left( \\frac{1}{F_1} - \\frac{1}{F_3} \\right) + \\left( \\frac{1}{F_2} - \\frac{1}{F_4} \\right) + \\left( \\frac{1}{F_3} - \\frac{1}{F_5} \\right) + \\dotsb \\\\\n&= \\frac{1}{F_1} + \\frac{1}{F_2} \\\\\n&= \\boxed{2}.\n\\end{align*}",
 "erwartet": "2",
 "level": 4,
 "antwort": "Wir nutzen die Fibonacci-Relation \\(F_{n+2}=F_{n+1}+F_n\\). Dann gilt\n\n\\[\n\\frac{F_{n+1}}{F_nF_{n+2}}\n=\n\\frac{F_{n+2}-F_n}{F_nF_{n+2}}\n=\n\\frac{1}{F_n}-\\frac{1}{F_{n+2}}.\n\\]\n\nDamit ist die Summe teleskopierend:\n\n\\[\n\\sum_{n=1}^N \\frac{F_{n+1}}{F_nF_{n+2}}\n=\n\\sum_{n=1}^N \\left(\\frac{1}{F_n}-\\frac{1}{F_{n+2}}\\right).\n\\]\n\nDie meisten Terme heben sich auf:\n\n\\[\n\\sum_{n=1}^N \\left(\\frac{1}{F_n}-\\frac{1}{F_{n+2}}\\right)\n=\n\\frac{1}{F_1}+\\frac{1}{F_2}\n-\n\\frac{1}{F_{N+1}}-\\frac{1}{F_{N+2}}.\n\\]\n\nDa \\(F_1=F_2=1\\), folgt\n\n\\[\n\\sum_{n=1}^N \\frac{F_{n+1}}{F_nF_{n+2}}\n=\n2-\\frac{1}{F_{N+1}}-\\frac{1}{F_{N+2}}.\n\\]\n\nFür \\(N\\to\\infty\\) gehen \\(\\frac{1}{F_{N+1}}\\) und \\(\\frac{1}{F_{N+2}}\\) gegen \\(0\\). Also ist\n\n\\[\n\\sum_{n=1}^\\infty \\frac{F_{n+1}}{F_nF_{n+2}}=2.\n\\]\n\n\\[\n\\boxed{2}\n\\]",
 "denken": "We need answer in German? User: \"Compute ... Löse die Aufgabe. Schreibe die endgültige Antwort ganz zum Schluss in der Form \\boxed{ANTWORT}.\" So likely respond in German. Need solve sum_{n=1}^∞ F_{n+1}/(F_n F_{n+2}). Need final boxed.\n\nWe need compute. Let's analyze.\n\nFibonacci: F0=0, F1=1, F2=1, F3=2, F4=3, F5=5...\nSum n=1: F2/(F1 F3)=1/(1*2)=1/2.\nn=2: F3/(F2 F4)=2/(1*3)=2/3.\nn=3: F4/(F3 F5)=3/(2*5)=3/10.\nn=4: 5/(3*8)=5/24.\nSum approx 0.5+0.6667+0.3+0.2083+... maybe converges? Terms ~ phi? F_{n+1}/(F_n F_{n+2}) ~ phi^n/(phi^n phi^{n+2})? Actually F_n ~ phi^n/sqrt5, term ~ phi^{n+1}/(phi^n phi^{n+2}) = 1/phi^3? Wait denominator ~ phi^{2n+2}/5, numerator ~ phi^{n+1}/sqrt5 => term ~ 5/(sqrt5 phi^{n+1})? decays phi^{-n}. converges.\n\nNeed exact sum.\n\nTry telescoping. Identity: 1/F_n - 1/F_{n+2}? Compute:\n1/F_n - 1/F_{n+2} = (F_{n+2}-F_n)/(F_n F_{n+2}) = F_{n+1}/(F_n F_{n+2}). Exactly! Since F_{n+2}=F_{n+1}+F_n, difference = F_{n+1}. So term = 1/F_n - 1/F_{n+2}.\n\nThen sum n=1 to N = (1/F1 - 1/F3) + (1/F2 - 1/F4)+... telescopes partially. Let's write:\nS_N = sum_{n=1}^N (1/F_n - 1/F_{n+2}) = sum_{n=1}^N 1/F_n - sum_{n=3}^{N+2} 1/F_n = 1/F1 + 1/F2 - 1/F_{N+1} - 1/F_{N+2}. Since F1=F2=1. So S_N = 2 - 1/F_{N+1} - 1/F_{N+2}. Limit N→∞: 2.\n\nCheck partial: N=1: 2 - 1/F2 - 1/F3 = 2 -1 -1/2=1/2. N=2: 2 -1/F3 -1/F4=2 -1/2 -1/3=7/6? But first two terms 1/2+2/3=7/6 yes. Limit 2.\n\nNeed final answer boxed{2}. Need maybe include German solution. Ensure final answer at end exactly \\boxed{2}. Could include details.\n\nDouble-check no issue n=1 F1=1 F2=1 F3=2. Yes.\n\nNeed final in German. \"Löse die Aufgabe.\" We can present.\n\nNeed ensure final answer ganz zum Schluss in form boxed. So last line \\boxed{2}. No extra after.\n\nLet's craft concise but enough. Use identity.\n",
 "gegeben": "2",
 "tokens": 1261
}