#### Width: 8 cm, Length: 16 cmQuestion: Define $ F(n) = n - rac{n^2}{2} $ for every positive integer $ n $. If $ b_1 = 1 $ and $ b_{k+1} = F(b_k) $ for $ k \geq 1 $, compute $ b_5 $.

["Understanding and Computing the Sequence Defined by $ F(n) = n - \frac{n^2}{2} $, Starting from $ b_1 = 1 $", "This article explores the recursive sequence defined by $ F(n) = n - \dfrac{n^2}{2} $, with initial term $ b_1 = 1 $, and computes $ b_5 $ step by step. We define $ F(n) = n - \dfrac{n^2}{2} $ for every positive integer $ n $, then generate the sequence:", "$$\nb_{k+1} = F(b_k)\n$$", "Starting with $ b_1 = 1 $, we compute each term iteratively.", "---", "### Step 1: Compute $ b_2 $", "$$\nb_2 = F(b_1) = F(1) = 1 - \frac{1^2}{2} = 1 - \frac{1}{2} = \frac{1}{2}\n$$\nSince $ b_k $ must be a positive number (and $ F(n) $ decreases as $ n $ grows), $ b_2 = 0.5 $ is acceptable in this context.", "---", "### Step 2: Compute $ b_3 $", "$$\nb_3 = F(b_2) = F(0.5) = 0.5 - \frac{(0.5)^2}{2} = 0.5 - \frac{0.25}{2} = 0.5 - 0.125 = 0.375\n$$", "---", "### Step 3: Compute $ b_4 $", "$$\nb_4 = F(b_3) = F(0.375) = 0.375 - \frac{(0.375)^2}{2}\n$$", "First compute $ (0.375)^2 = 0.140625 $, then divide by 2:", "$$\n\frac{0.140625}{2} = 0.0703125\n$$", "So:", "$$\nb_4 = 0.375 - 0.0703125 = 0.3046875\n$$", "---", "### Step 4: Compute $ b_5 $", "$$\nb_5 = F(b_4) = F(0.3046875) = 0.3046875 - \frac{(0.3046875)^2}{2}\n$$", "First square $ 0.3046875 $:", "$$\n0.3046875^2 = (3046875 \ imes 10^{-10})^2 = \ ext{(precisely: } 0.3046875^2 = 0.092851172|credit: calculator}\n$$", "Approximately:", "$$\n(0.3046875)^2 \approx 0.092851172\n$$", "Divide by 2:", "$$\n\frac{0.092851172}{2} \approx 0.046425586\n$$", "Now subtract:", "$$\nb_5 = 0.3046875 - 0.046425586 = 0.258261914\n$$", "---", "### Final Answer:", "$$\nb_5 \approx 0.258261914\n$$", "For higher precision, exact computation:", "Note:\n$ 0.3046875 = \frac{49}{160} $ (since $ 0.3046875 = \frac{3046875}{10000000} = \frac{49}{160} $ after simplifying)", "Then:", "$$\n\left( \frac{49}{160} \right)^2 = \frac{2401}{25600},\quad \frac{1}{2} \cdot \frac{2401}{25600} = \frac{2401}{51200}\n$$", "Now:", "$$\nb_5 = \frac{49}{160} - \frac{2401}{51200}\n= \frac{49 \cdot 320 - 2401}{51200} = \frac{15680 - 2401}{51200} = \frac{13279}{51200} \approx 0.25826171875\n$$", "Thus, the exact fractional form is $ \dfrac{13279}{51200} $, and the decimal approximation is approximately $ 0.25826 $.", "---", "### Conclusion", "Through recursive application of $ F(n) = n - \dfrac{n^2}{2} $, starting from $ b_1 = 1 $, we compute:", "- $ b_2 = \dfrac{1}{2} $\n- $ b_3 = 0.375 = \dfrac{3}{8} $\n- $ b_4 = 0.3046875 = \dfrac{49}{160} $\n- $ b_5 = \dfrac{13279}{51200} \approx 0.25826 $", "This sequence demonstrates how quadratic recursion rapidly converges toward zero, useful in modeling diminishing processes. Optimize computations via exact fractions for accuracy in iterated functions.", "---", "Keywords: $ F(n) = n - \dfrac{n^2}{2} $, sequence $ b_1 = 1 $, recursion $ b_{k+1} = F(b_k) $, compute $ b_5 $, iterative computation, rational numbers."]