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Difficulty: Very hardEstimation, Rounding, and Sequences

A sequence a1,a2,a3,…a_1, a_2, a_3, \dots is defined by a1=3a_1 = 3 and an+1=an+12a_{n+1} = \sqrt{a_n + 12} for all integers n≥1n \geq 1. If S100=∑k=1100akS_{100} = \sum_{k=1}^{100} a_k, what is the value of S100S_{100} rounded to the nearest integer?

  1. A
    396
  2. B
    398
  3. 399Answer
  4. D
    400
  5. E
    401

Answer

399
The sequence terms approach the fixed limit L=4L = 4 defined by L=L+12L = \sqrt{L + 12}. Each term aka_k is strictly less than 4, with difference ϵk=4−ak\epsilon_k = 4 - a_k. Evaluating the first few differences yields ϵ1=1\epsilon_1 = 1, ϵ2≈0.127\epsilon_2 \approx 0.127, ϵ3≈0.016\epsilon_3 \approx 0.016, ϵ4≈0.002\epsilon_4 \approx 0.002, and subsequent terms are negligible. The total sum S100=100×4−∑k=1100ϵk≈400−1.145=398.855S_{100} = 100 \times 4 - \sum_{k=1}^{100} \epsilon_k \approx 400 - 1.145 = 398.855. Rounding 398.855398.855 to the nearest integer gives 399.

Step-by-Step Solution

1
Determine the limit/fixed point of the recursive sequence.
Solving L=L+12L = \sqrt{L + 12} yields L2−L−12=0  ⟹  (L−4)(L+3)=0L^2 - L - 12 = 0 \implies (L - 4)(L + 3) = 0. Since terms are positive, L=4L = 4.
Understanding the upper limit L=4L = 4 helps set up a upper bound for all terms in the sequence.
2
Calculate the first few terms and their deficits ϵk=4−ak\epsilon_k = 4 - a_k from the limit 4.
ϵ1=4−3=1\epsilon_1 = 4 - 3 = 1; ϵ2=4−15≈0.12702\epsilon_2 = 4 - \sqrt{15} \approx 0.12702; ϵ3=4−15.87298≈0.01591\epsilon_3 = 4 - \sqrt{15.87298} \approx 0.01591; ϵ4≈0.00199\epsilon_4 \approx 0.00199; ϵ5≈0.00025\epsilon_5 \approx 0.00025.
Because convergence is extremely fast (each error term shrinks by a factor greater than 8), only the first few deficit terms contribute meaningfully to the sum.
3
Sum the total deficit ∑k=1100ϵk\sum_{k=1}^{100} \epsilon_k.
Total deficit ≈1+0.12702+0.01591+0.00199+0.00025+⋯≈1.145\approx 1 + 0.12702 + 0.01591 + 0.00199 + 0.00025 + \dots \approx 1.145.
Adding all error terms gives the precise deviation of S100S_{100} from 100×4=400100 \times 4 = 400.
4
Compute S100S_{100} and round to the nearest integer.
S100=400−1.145=398.855S_{100} = 400 - 1.145 = 398.855. Rounded to the nearest integer, 398.855398.855 rounds to 399399.
Since 398.855≥398.5398.855 \geq 398.5, rounding to the nearest integer yields 399.

Key Concept

Fixed-point sequence convergence, summation bounding, and rounding
Estimated Time:2m 30s
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Topics

Estimation, Rounding, and Sequences

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