Estimation, Rounding, and Sequences

12 questions

Question 1Question

A sequence a1,a2,a3,a_1, a_2, a_3, \dots is defined by an=1n+n+2a_n = \frac{1}{\sqrt{n} + \sqrt{n+2}} for all positive integers nn. If S=n=198anS = \sum_{n=1}^{98} a_n, which of the following is closest to the value of SS when rounded to the nearest tenth?

Show answer & explanation

Answer: 8.8

Answer

The value of SS rounded to the nearest tenth is 8.8.
The term ana_n simplifies to n+2n2\frac{\sqrt{n+2} - \sqrt{n}}{2} upon rationalizing the denominator. Summing from n=1n=1 to 9898 causes all middle terms to cancel out, leaving 99+100122\frac{\sqrt{99} + \sqrt{100} - \sqrt{1} - \sqrt{2}}{2}. Substituting 100=10\sqrt{100} = 10, 1=1\sqrt{1} = 1, 999.95\sqrt{99} \approx 9.95, and 21.41\sqrt{2} \approx 1.41 gives approximately 8.778.77, which rounds to 8.8.

Step-by-Step Solution

1
Rationalize the general term ana_n
an=1n+n+2n+2nn+2n=n+2n(n+2)n=n+2n2a_n = \frac{1}{\sqrt{n} + \sqrt{n+2}} \cdot \frac{\sqrt{n+2} - \sqrt{n}}{\sqrt{n+2} - \sqrt{n}} = \frac{\sqrt{n+2} - \sqrt{n}}{(n+2) - n} = \frac{\sqrt{n+2} - \sqrt{n}}{2}
Eliminating radicals from the denominator reveals the underlying telescoping structure of the sequence.
2
Expand the summation S=n=198anS = \sum_{n=1}^{98} a_n
S=12[(31)+(42)+(53)++(9997)+(10098)]S = \frac{1}{2} \left[ (\sqrt{3} - \sqrt{1}) + (\sqrt{4} - \sqrt{2}) + (\sqrt{5} - \sqrt{3}) + \dots + (\sqrt{99} - \sqrt{97}) + (\sqrt{100} - \sqrt{98}) \right]
Writing out initial and final terms demonstrates which terms cancel.
3
Simplify the telescoping sum
S=99+100122=99+10122=9+9922S = \frac{\sqrt{99} + \sqrt{100} - \sqrt{1} - \sqrt{2}}{2} = \frac{\sqrt{99} + 10 - 1 - \sqrt{2}}{2} = \frac{9 + \sqrt{99} - \sqrt{2}}{2}
All intermediate terms cancel out, leaving two positive boundary terms and two negative boundary terms.
4
Estimate square root values and perform rounding
Since 999.94987\sqrt{99} \approx 9.94987 and 21.41421\sqrt{2} \approx 1.41421, S9+9.949871.414212=17.535662=8.767838.8S \approx \frac{9 + 9.94987 - 1.41421}{2} = \frac{17.53566}{2} = 8.76783 \approx 8.8
Evaluating the radicals to two decimal places allows accurate rounding to the nearest tenth.

Key Concept

Telescoping Series Summation and Square Root Estimation
Question 2Question

An index value I0=1000I_0 = 1{}000 increases over a 6-month period. For each month nn from 1 to 6, the index value InI_n is calculated by increasing the previous month's value In1I_{n-1} by 5%5\% and then rounding the result to the nearest integer (with half-integers rounded up). Let U6=1000×(1.05)6U_6 = 1{}000 \times (1.05)^6 represent the exact unrounded compounded value at month 6.

Which of the following statements must be true? Select all such statements.

Select all that apply

Show answer & explanation

Answer: I2=1103I_2 = 1{}103; The sequence of monthly increments InIn1I_n - I_{n-1} for n=1,2,,6n = 1, 2, \dots, 6 is strictly increasing.

Answer

The statement specifying that I2=1103I_2 = 1{}103 and the statement asserting that the sequence of monthly increments InIn1I_n - I_{n-1} is strictly increasing are both correct.
The statement giving I2=1103I_2 = 1{}103 is correct because 1050×1.05=1102.51{}050 \times 1.05 = 1{}102.5, which rounds up to 11031{}103. The statement regarding the sequence of monthly increments is correct because the increments 50,53,55,58,61,6450, 53, 55, 58, 61, 64 strictly increase.

Step-by-Step Solution

1
Calculate each term of the sequence InI_n by applying a 5% increase and rounding to the nearest integer.
I0=1000I_0 = 1{}000; I1=round(1000×1.05)=1050I_1 = \text{round}(1{}000 \times 1.05) = 1{}050; I2=round(1050×1.05)=round(1102.5)=1103I_2 = \text{round}(1{}050 \times 1.05) = \text{round}(1{}102.5) = 1{}103; I3=round(1103×1.05)=round(1158.15)=1158I_3 = \text{round}(1{}103 \times 1.05) = \text{round}(1{}158.15) = 1{}158; I4=round(1158×1.05)=round(1215.9)=1216I_4 = \text{round}(1{}158 \times 1.05) = \text{round}(1{}215.9) = 1{}216; I5=round(1216×1.05)=round(1276.8)=1277I_5 = \text{round}(1{}216 \times 1.05) = \text{round}(1{}276.8) = 1{}277; I6=round(1277×1.05)=round(1340.85)=1341I_6 = \text{round}(1{}277 \times 1.05) = \text{round}(1{}340.85) = 1{}341.
This establishes the exact sequence of rounded monthly values.
2
Evaluate the statement that I2=1103I_2 = 1{}103.
From Step 1, I2=1103I_2 = 1{}103.
This directly confirms the validity of the first statement.
3
Compute the sequence of monthly increments InIn1I_n - I_{n-1} for n=1,2,,6n = 1, 2, \dots, 6.
Increments: I1I0=50I_1 - I_0 = 50, I2I1=53I_2 - I_1 = 53, I3I2=55I_3 - I_2 = 55, I4I3=58I_4 - I_3 = 58, I5I4=61I_5 - I_4 = 61, I6I5=64I_6 - I_5 = 64.
Since 50<53<55<58<61<6450 < 53 < 55 < 58 < 61 < 64, the sequence of increments is strictly increasing.
4
Compare I6I_6 with U6=1000×(1.05)6U_6 = 1{}000 \times (1.05)^6.
U6=1000×1.3400956...1340.10U_6 = 1{}000 \times 1.3400956... \approx 1{}340.10. Since I6=1341I_6 = 1{}341, I6>U6I_6 > U_6.
The statement claiming U6>I6U_6 > I_6 is false.

Key Concept

Error propagation in recursive sequence rounding and non-distributivity of exponents over sums
Estimated Time:2m 30s
Question 3Question

A financial forecasting model predicts monthly sales revenue using a sequence where the initial revenue in month 1 is R1=$80000R_1 = \$80{}000, and the revenue for each subsequent month nn (for n2n \ge 2) is calculated using the recursive formula Rn=0.75Rn1+5000R_n = 0.75 R_{n-1} + 5{}000. If each monthly revenue figure RnR_n is rounded to the nearest hundred dollars before summing, what is the estimated total revenue, in dollars, for the first 4 months combined?

Show answer & explanation

Answer: 244100

Answer

The estimated total revenue for the first 4 months combined is 244,100 dollars.
By evaluating the recursive formula Rn=0.75Rn1+5000R_n = 0.75 R_{n-1} + 5{}000 for months 1 through 4, we obtain R1=80000R_1 = 80{}000, R2=65000R_2 = 65{}000, R3=53750R_3 = 53{}750, and R4=45312.50R_4 = 45{}312.50. Rounding each to the nearest hundred yields 8000080{}000, 6500065{}000, 5380053{}800, and 4530045{}300, respectively. Summing these four rounded figures gives 244100244{}100.

Step-by-Step Solution

1
Find the first term R1R_1 and its rounded value.
R1=80000R_1 = 80{}000, which rounds to 8000080{}000.
Given initial term.
2
Calculate the second term R2R_2 using the recursive formula and round it.
R2=0.75(80000)+5000=65000R_2 = 0.75(80{}000) + 5{}000 = 65{}000, which rounds to 6500065{}000.
Apply R2=0.75R1+5000R_2 = 0.75 R_1 + 5{}000.
3
Calculate the third term R3R_3 using unrounded R2R_2 and round to the nearest hundred.
R3=0.75(65000)+5000=53750R_3 = 0.75(65{}000) + 5{}000 = 53{}750, which rounds to 5380053{}800.
Apply R3=0.75R2+5000R_3 = 0.75 R_2 + 5{}000 and round to the nearest 100.
4
Calculate the fourth term R4R_4 using unrounded R3R_3 and round to the nearest hundred.
R4=0.75(53750)+5000=45312.50R_4 = 0.75(53{}750) + 5{}000 = 45{}312.50, which rounds to 4530045{}300.
Apply R4=0.75R3+5000R_4 = 0.75 R_3 + 5{}000 and round to the nearest 100.
5
Add the four rounded terms together.
80000+65000+53800+45300=24410080{}000 + 65{}000 + 53{}800 + 45{}300 = 244{}100.
Compute total estimated revenue as specified.

Key Concept

Estimation, Rounding, and Sequences
Question 4Question

A sequence a1,a2,a3,,ana_1, a_2, a_3, \dots, a_n is defined by a1=1a_1 = 1 and an=2an1+3a_n = 2a_{n-1} + 3 for all integers n2n \geq 2. What is the value of a4a_4?

Show answer & explanation

Answer: 2929

Answer

The value of a4a_4 is 2929.
Using the recursive rule an=2an1+3a_n = 2a_{n-1} + 3 starting with a1=1a_1 = 1, we compute a2=2(1)+3=5a_2 = 2(1) + 3 = 5, a3=2(5)+3=13a_3 = 2(5) + 3 = 13, and finally a4=2(13)+3=29a_4 = 2(13) + 3 = 29.

Step-by-Step Solution

1
Calculate the second term a2a_2 using the recursive formula with a1=1a_1 = 1.
a2=2(1)+3=5a_2 = 2(1) + 3 = 5
The recursive rule requires the previous term to compute the next term.
2
Calculate the third term a3a_3 using a2=5a_2 = 5.
a3=2(5)+3=13a_3 = 2(5) + 3 = 13
Apply the recursive definition an=2an1+3a_n = 2a_{n-1} + 3 for n=3n = 3.
3
Calculate the fourth term a4a_4 using a3=13a_3 = 13.
a4=2(13)+3=29a_4 = 2(13) + 3 = 29
Apply the recursive definition an=2an1+3a_n = 2a_{n-1} + 3 for n=4n = 4.

Key Concept

Recursive Sequences
Question 5Question

A sequence of numbers begins with t1=4t_1 = 4. For all integers n2n \geq 2, each term is defined by tn=3tn12t_n = 3t_{n-1} - 2. What is the value of t5t_5?

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Answer: 244

Answer

The 5th term of the sequence, t5t_5, is equal to 244.
Applying the recursive relation tn=3tn12t_n = 3t_{n-1} - 2 step-by-step starting from t1=4t_1 = 4 gives t2=10t_2 = 10, t3=28t_3 = 28, t4=82t_4 = 82, and finally t5=244t_5 = 244.

Step-by-Step Solution

1
Find the second term, t2t_2, using t1=4t_1 = 4.
t2=3(4)2=122=10t_2 = 3(4) - 2 = 12 - 2 = 10
Substitute n=2n = 2 into the recursive rule tn=3tn12t_n = 3t_{n-1} - 2.
2
Find the third term, t3t_3, using t2=10t_2 = 10.
t3=3(10)2=302=28t_3 = 3(10) - 2 = 30 - 2 = 28
Substitute n=3n = 3 into the recursive rule.
3
Find the fourth term, t4t_4, using t3=28t_3 = 28.
t4=3(28)2=842=82t_4 = 3(28) - 2 = 84 - 2 = 82
Substitute n=4n = 4 into the recursive rule.
4
Find the fifth term, t5t_5, using t4=82t_4 = 82.
t5=3(82)2=2462=244t_5 = 3(82) - 2 = 246 - 2 = 244
Substitute n=5n = 5 into the recursive rule.

Key Concept

Evaluating terms in a recursively defined sequence
Question 6Question

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 n1n \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?

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Answer: 399

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=4ak\epsilon_k = 4 - a_k. Evaluating the first few differences yields ϵ1=1\epsilon_1 = 1, ϵ20.127\epsilon_2 \approx 0.127, ϵ30.016\epsilon_3 \approx 0.016, ϵ40.002\epsilon_4 \approx 0.002, and subsequent terms are negligible. The total sum S100=100×4k=1100ϵk4001.145=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 L2L12=0    (L4)(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=4ak\epsilon_k = 4 - a_k from the limit 4.
ϵ1=43=1\epsilon_1 = 4 - 3 = 1; ϵ2=4150.12702\epsilon_2 = 4 - \sqrt{15} \approx 0.12702; ϵ3=415.872980.01591\epsilon_3 = 4 - \sqrt{15.87298} \approx 0.01591; ϵ40.00199\epsilon_4 \approx 0.00199; ϵ50.00025\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=4001.145=398.855S_{100} = 400 - 1.145 = 398.855. Rounded to the nearest integer, 398.855398.855 rounds to 399399.
Since 398.855398.5398.855 \geq 398.5, rounding to the nearest integer yields 399.

Key Concept

Fixed-point sequence convergence, summation bounding, and rounding
Estimated Time:2m 30s
Question 7Question

A sequence of positive real numbers a1,a2,a3,a_1, a_2, a_3, \dots is defined by a1=3a_1 = 3 and an+1=an+2ana_{n+1} = a_n + \frac{2}{a_n} for all integers n1n \ge 1. Which of the following statements must be true? Select all that apply.

Select all that apply

Show answer & explanation

Answer: The sequence a1,a2,a3,a_1, a_2, a_3, \dots is strictly increasing.; The term a50a_{50} is strictly greater than 1414.; The sequence of consecutive term differences dn=an+1and_n = a_{n+1} - a_n is strictly decreasing for n1n \ge 1.

Answer

The correct statements are that the sequence is strictly increasing, the term a50a_{50} is strictly greater than 14, and the sequence of consecutive term differences dn=an+1and_n = a_{n+1} - a_n is strictly decreasing.
The statement asserting strict monotonicity is correct because an+1an=2an>0a_{n+1} - a_n = \frac{2}{a_n} > 0 for positive terms. The statement regarding a50>14a_{50} > 14 is correct because a502>32+4(49)=205>196a_{50}^2 > 3^2 + 4(49) = 205 > 196. The statement concerning consecutive differences is correct because 2an\frac{2}{a_n} strictly decreases as ana_n increases.

Step-by-Step Solution

1
Analyze monotonicity of the sequence
an+1an=2an>0a_{n+1} - a_n = \frac{2}{a_n} > 0 for all n1n \ge 1
Since a1=3>0a_1 = 3 > 0, all terms remain positive, making each term strictly larger than the previous.
2
Analyze the sequence of term differences
dn=2and_n = \frac{2}{a_n} decreases as ana_n increases
Since ana_n grows strictly monotonically, its reciprocal strictly decreases, so dn+1<dnd_{n+1} < d_n.
3
Derive a lower bound for a50a_{50} using quadratic expansion
a502>205    a50>14a_{50}^2 > 205 \implies a_{50} > 14
Expanding ak+12=ak2+4+4ak2>ak2+4a_{k+1}^2 = a_k^2 + 4 + \frac{4}{a_k^2} > a_k^2 + 4 and summing from k=1k=1 to 4949 gives a502>32+4(49)=205>196=142a_{50}^2 > 3^2 + 4(49) = 205 > 196 = 14^2.
4
Derive an upper bound for a50a_{50} to test the rounding statement
a50<14.45a_{50} < 14.45, so rounding to the nearest integer yields 14
Using 4ak2<44k+5\frac{4}{a_k^2} < \frac{4}{4k+5}, the sum of error terms is bounded above by 04944x+5dx=ln(40.2)3.7\int_0^{49} \frac{4}{4x+5} dx = \ln(40.2) \approx 3.7. Thus a502<208.7<14.52a_{50}^2 < 208.7 < 14.5^2, so a50a_{50} rounds to 14, not 15.
5
Estimate a100a_{100} to test the magnitude bound
a100<409.420.23<25a_{100} < \sqrt{409.4} \approx 20.23 < 25
Summing the squared recurrence up to 100 terms gives a1002<405+ln(80.2)409.4a_{100}^2 < 405 + \ln(80.2) \approx 409.4, showing a100a_{100} cannot exceed 25.

Key Concept

Estimation of non-linear recursive sequences using squared bounds and integral comparison.
Estimated Time:3m 0s
Question 8Question

A sequence of 50 numerical measurements x1,x2,,x50x_1, x_2, \dots, x_{50} is collected. Each measurement xix_i is rounded to the nearest tenth to produce a rounded value rir_i. The sum of the 50 rounded values, i=150ri\sum_{i=1}^{50} r_i, is equal to 250.0250.0. If SS represents the true sum of the unrounded measurements i=150xi\sum_{i=1}^{50} x_i, what is the maximum possible percent error of the rounded sum relative to the true sum SS, rounded to the nearest hundredth of a percent?

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Answer: 1.01%1.01\%

Answer

The maximum possible percent error of the rounded sum relative to the true sum is 1.01%1.01\%.
When rounding numbers to the nearest tenth, the maximum error for each number is 0.050.05. For 50 numbers, the maximum possible error in the sum is 50×0.05=2.550 \times 0.05 = 2.5. The true sum SS therefore lies in the range [247.5,252.5][247.5, 252.5]. To maximize the percent error relative to SS, defined as 250.0SS×100%\frac{|250.0 - S|}{S} \times 100\%, we use the maximum numerator 2.52.5 and the smallest possible denominator S=247.5S = 247.5. This yields 2.5247.5×100%1.01%\frac{2.5}{247.5} \times 100\% \approx 1.01\%.

Step-by-Step Solution

1
Determine the maximum rounding error for a single term
Maximum error per measurement is xiri0.05|x_i - r_i| \le 0.05
When rounding to the nearest tenth, any value within 0.050.05 of the rounded value rounds to that tenth.
2
Calculate the maximum cumulative error for the sequence sum
Maximum total error =50×0.05=2.5= 50 \times 0.05 = 2.5
The maximum difference between the true sum SS and the rounded sum 250.0250.0 occurs when all individual rounding errors accumulate in the same direction.
3
Find the range of possible true sum values SS
247.5S252.5247.5 \le S \le 252.5
Subtracting and adding the maximum error of 2.52.5 from the rounded sum 250.0250.0 establishes the bounds for SS.
4
Set up and maximize the percent error expression
Max percent error occurs at minimum S=247.5S = 247.5, giving 2.5247.5×100%1.0101%\frac{2.5}{247.5} \times 100\% \approx 1.0101\%
Percent error relative to SS is given by 250.0SS×100%\frac{|250.0 - S|}{S} \times 100\%. To maximize this ratio, we divide the maximum numerator 2.52.5 by the smallest positive denominator S=247.5S = 247.5.

Key Concept

Error propagation in sequence sums and optimizing percent error base values
Question 9Question

A sequence of 40 numbers a1,a2,,a40a_1, a_2, \dots, a_{40} is defined by the formula ak=15+3k8a_k = 15 + \frac{3k}{8} for each integer kk from 1 to 40. Each term aka_k is rounded to the nearest integer to form a new sequence b1,b2,,b40b_1, b_2, \dots, b_{40}. (Note: numbers ending in .5.5 are rounded up to the next integer.) What is the value of k=140bkk=140ak\sum_{k=1}^{40} b_k - \sum_{k=1}^{40} a_k?

Show answer & explanation

Answer: 2.52.5

Answer

The sum of the rounded sequence exceeds the sum of the exact sequence by 2.52.5.
Evaluating ak=15+3k8a_k = 15 + \frac{3k}{8} for k=1,2,,8k = 1, 2, \dots, 8 yields fractional parts of 0.375,0.75,0.125,0.5,0.875,0.25,0.625,0.375, 0.75, 0.125, 0.5, 0.875, 0.25, 0.625, and 0.00.0. The individual rounding errors (bkak)(b_k - a_k) for these terms are 0.375,+0.25,0.125,+0.5,+0.125,0.25,+0.375,-0.375, +0.25, -0.125, +0.5, +0.125, -0.25, +0.375, and 0.00.0. Summing these errors yields +0.5+0.5 per 8-term period. For 40 terms (5 full periods), the total error sum is 5×0.5=2.55 \times 0.5 = 2.5.

Step-by-Step Solution

1
Analyze the fractional part of ak=15+3k8a_k = 15 + \frac{3k}{8} over one complete 8-term period.
The fractional parts for k=1,2,,8k = 1, 2, \dots, 8 are 0.375,0.75,0.125,0.5,0.875,0.25,0.625,0.00.375, 0.75, 0.125, 0.5, 0.875, 0.25, 0.625, 0.0 respectively.
Since 3k8\frac{3k}{8} repeats its fractional pattern modulo 8, examining one period reveals the periodic rounding behavior.
2
Calculate the rounding error (bkak)(b_k - a_k) for each term in the 8-term cycle.
The differences (bkak)(b_k - a_k) for k=1k=1 to 88 are: 0.375,+0.25,0.125,+0.5,+0.125,0.25,+0.375,0.0-0.375, +0.25, -0.125, +0.5, +0.125, -0.25, +0.375, 0.0.
Rounding to the nearest integer shifts each number by its distance to that integer. Half-integers like 0.50.5 round up, yielding a +0.5+0.5 difference.
3
Sum the rounding errors over one 8-term cycle.
(0.375+0.375)+(0.250.25)+(0.125+0.125)+0.5+0.0=+0.5(-0.375 + 0.375) + (0.25 - 0.25) + (-0.125 + 0.125) + 0.5 + 0.0 = +0.5.
Symmetric fractional pairs cancel out, leaving only the +0.5+0.5 error from the half-integer term.
4
Multiply the single-cycle error sum by the number of full cycles in 40 terms.
Since 40=5×840 = 5 \times 8, total difference =5×0.5=2.5= 5 \times 0.5 = 2.5.
The 40 terms consist of 5 identical 8-term periodic cycles.

Key Concept

Periodic error analysis in sequence rounding and summation
Question 10Question

A sequence of 24 positive numbers a1,a2,,a24a_1, a_2, \dots, a_{24} is defined by an=120n(n+1)a_n = \frac{120}{n(n+1)} for each integer nn from 11 to 2424. Let T=n=124anT = \sum_{n=1}^{24} a_n be the exact sum of all 24 terms. Which of the following statements must be true? Select all that apply.

Select all that apply

Show answer & explanation

Answer: The value of TT rounded to the nearest integer is 115115.; The sum of the first 4 terms, n=14an\sum_{n=1}^{4} a_n, represents more than 80%80\% of the total sum TT.

Answer

The statements confirming that TT rounded to the nearest integer is 115115, and that the sum of the first 4 terms represents more than 80%80\% of TT, are both correct.
The exact sum of the telescoping sequence simplifies to T=120(1125)=115.2T = 120 \left(1 - \frac{1}{25}\right) = 115.2. Rounding 115.2115.2 to the nearest integer gives 115115. Furthermore, the partial sum of the first four terms is 120(115)=96120 \left(1 - \frac{1}{5}\right) = 96, which accounts for 96115.2=5683.33%\frac{96}{115.2} = \frac{5}{6} \approx 83.33\% of the total sum, exceeding 80%80\%.

Step-by-Step Solution

1
Decompose the sequence formula using partial fractions.
an=120n(n+1)=120(1n1n+1)a_n = \frac{120}{n(n+1)} = 120 \left( \frac{1}{n} - \frac{1}{n+1} \right).
Rewriting the terms as partial fractions converts the sum into a telescoping series.
2
Calculate the exact total sum TT.
T=120[(112)+(1213)++(124125)]=120(1125)=120×0.96=115.2T = 120 \left[ \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{24} - \frac{1}{25}\right) \right] = 120 \left(1 - \frac{1}{25}\right) = 120 \times 0.96 = 115.2.
All intermediate terms cancel out, leaving only the first and last components.
3
Evaluate the statement regarding rounding TT to the nearest integer.
115.2115.2 rounded to the nearest integer is 115115, making the rounding statement true.
Since the decimal part .2.2 is less than .5.5, the number rounds down to 115115.
4
Evaluate the sum of the first 4 terms and compare its percentage to TT.
n=14an=120(115)=96\sum_{n=1}^{4} a_n = 120 \left(1 - \frac{1}{5}\right) = 96. The percentage is 96115.2=5683.33%>80%\frac{96}{115.2} = \frac{5}{6} \approx 83.33\% > 80\%.
Comparing 83.33%83.33\% to 80%80\% confirms that the partial sum statement is true.

Key Concept

Telescoping Series Summation and Percent Estimation
Question 11Question

A sequence of numbers a1,a2,a3,a_1, a_2, a_3, \dots is defined by a1=5a_1 = 5 and an+1=an22a_{n+1} = a_n^2 - 2 for all positive integers n1n \ge 1. Which of the following is the value of the sum n=141an\sum_{n=1}^{4} \frac{1}{a_n}, rounded to the nearest hundredth?

Show answer & explanation

Answer: 0.25

Answer

0.25
Evaluating the recurrence relation yields a1=5a_1 = 5, a2=23a_2 = 23, a3=527a_3 = 527, and a4=277,727a_4 = 277,727. Summing their reciprocals produces 15+123+1527+1277,7270.20+0.043478+0.001898+0.000004=0.24538\frac{1}{5} + \frac{1}{23} + \frac{1}{527} + \frac{1}{277,727} \approx 0.20 + 0.043478 + 0.001898 + 0.000004 = 0.24538. Rounding 0.245380.24538 to the nearest hundredth yields 0.25.

Step-by-Step Solution

1
Calculate the first four terms of the defined sequence using the recurrence relation an+1=an22a_{n+1} = a_n^2 - 2.
a1=5a_1 = 5, a2=522=23a_2 = 5^2 - 2 = 23, a3=2322=527a_3 = 23^2 - 2 = 527, and a4=52722=277,727a_4 = 527^2 - 2 = 277,727.
The recurrence rule determines each subsequent term from the preceding term.
2
Compute the sum of reciprocals n=141an=15+123+1527+1277,727\sum_{n=1}^{4} \frac{1}{a_n} = \frac{1}{5} + \frac{1}{23} + \frac{1}{527} + \frac{1}{277,727}.
15=0.2\frac{1}{5} = 0.2, 1230.043478\frac{1}{23} \approx 0.043478, 15270.001898\frac{1}{527} \approx 0.001898, and 1277,7270.0000036\frac{1}{277,727} \approx 0.0000036. Sum 0.24538\approx 0.24538.
Converting each fraction term to decimal form allows straightforward addition.
3
Round the calculated sum 0.245380.24538 to the nearest hundredth.
Since the thousandths digit is 55, 0.245380.24538 rounds up to 0.250.25.
Standard rounding rules dictate rounding up when the digit to the right of the target decimal place is 55 or greater.

Key Concept

Defined sequence terms evaluation, reciprocal sum estimation, and decimal rounding
Estimated Time:1m 30s
Question 12Question

A sequence of numbers a1,a2,a3,a_1, a_2, a_3, \dots is defined by a1=50a_1 = 50 and an+1=13an+15a_{n+1} = \frac{1}{3} a_n + 15 for all positive integers n1n \ge 1. Each term ana_n is rounded to the nearest integer to produce a secondary sequence bnb_n. What is the value of n=14bn\sum_{n=1}^{4} b_n?

Show answer & explanation

Answer: 132

Answer

132
Evaluating each term yields a1=50a_1 = 50, a2=31.6667...a_2 = 31.6667..., a3=25.5556...a_3 = 25.5556..., and a4=23.5185...a_4 = 23.5185.... Rounding each term individually to the nearest integer gives b1=50b_1 = 50, b2=32b_2 = 32, b3=26b_3 = 26, and b4=24b_4 = 24. Their sum is 50+32+26+24=13250 + 32 + 26 + 24 = 132.

Step-by-Step Solution

1
Calculate the exact value of the first term a1a_1 and its rounded value b1b_1.
a1=50a_1 = 50, which is an integer. Thus, b1=50b_1 = 50.
Given initial value.
2
Calculate the exact value of a2a_2 using the recursive formula a2=13a1+15a_2 = \frac{1}{3}a_1 + 15 and round to nearest integer.
a2=13(50)+15=503+15=31.6667...a_2 = \frac{1}{3}(50) + 15 = \frac{50}{3} + 15 = 31.6667.... Rounding to the nearest integer gives b2=32b_2 = 32.
Apply sequence relation and rounding rule.
3
Calculate the exact value of a3=13a2+15a_3 = \frac{1}{3}a_2 + 15 and round to nearest integer.
a3=13(953)+15=959+15=2309=25.5556...a_3 = \frac{1}{3}\left(\frac{95}{3}\right) + 15 = \frac{95}{9} + 15 = \frac{230}{9} = 25.5556.... Rounding to the nearest integer gives b3=26b_3 = 26.
Apply sequence relation and rounding rule.
4
Calculate the exact value of a4=13a3+15a_4 = \frac{1}{3}a_3 + 15 and round to nearest integer.
a4=13(2309)+15=23027+15=63527=23.5185...a_4 = \frac{1}{3}\left(\frac{230}{9}\right) + 15 = \frac{230}{27} + 15 = \frac{635}{27} = 23.5185.... Rounding to the nearest integer gives b4=24b_4 = 24.
Apply sequence relation and rounding rule.
5
Sum the four rounded terms b1+b2+b3+b4b_1 + b_2 + b_3 + b_4.
n=14bn=50+32+26+24=132\sum_{n=1}^{4} b_n = 50 + 32 + 26 + 24 = 132.
Add the individual rounded terms.

Key Concept

Recursive sequence evaluation and term-by-term rounding vs total sum rounding.