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(a)Write down, in descending powers of p, the first 3 terms in the binomial expansion of:
(2p + 3)7
Give each term in its simplest form. For example, the first term should be of the form ap7, where a is a constant.
(a)
Spiders build webs out of silk. This question involves two different models of how a spider might build its web. The functions in each model give lengths in centimetres.
(a)In the first model, the web is made in stages. First, the spider makes Stage 1 of the web. It then adds extra silk to Stage 1 to make Stage 2, and so on. The table shows S(n), the total length of silk in each stage, for the first three stages. It also shows A(n): for n > 1, A(n) is the length of extra silk needed to make Stage n from Stage n − 1. A(1) is set to be 15·8.
| Stage | 1 | 2 | 3 |
|---|---|---|---|
| S(n): total silk | 15·8 | 37·8 | 66 |
| A(n): extra silk | 15·8 |
(i)Use the values in the table to show that A(2) = 22 and find the value of A(3).
(a)(i)
The numbers A(1), A(2), A(3), … form an arithmetic sequence.
(ii)Find an expression in n for A(n), where n ∈ ℕ.
(a)(ii)
(iii)Hence, or otherwise, find the value of A(100).
(a)(iii)
(iv)S(n) is the total length of silk needed to make Stage n, where n ∈ ℕ. By using the formula for the sum of an arithmetic series, or otherwise, show that S(n) = 3·1n2 + 12·7n.
(a)(iv)
(v)Stage k is the first stage for which the total length of silk is more than 10 m, in this model. Using the expression for S(n) above, solve an equation to find the value of k. Remember that S(n) gives the total length of silk, in centimetres.
(a)(v)
(b)In the second model, the web is made in laps. The diagram shows the first lap. The line segments marked O1, O2, and O3 are the first 3 orbitals of the web. The lengths of the orbitals form a geometric sequence. The length of O1 is 0·5 cm and the length of O2 is 0·53 cm.
(i)Find the length of O3.
(b)(i)
(ii)Write an expression in n for the total length of the first n orbitals of the web, where n ∈ ℕ.
(b)(ii)
(iii)There are exactly 18 orbitals in each lap of the web. Write an expression in k for the total length of the first k laps, where k ∈ ℕ.
(b)(iii)
The first three patterns in a sequence of patterns are shown below. Each pattern is made up of dots (●) at points in the co-ordinate plane that have integer co-ordinates. Pattern 1 has a dot at all such points that are a distance of 1 unit from (0, 0), and Pattern 2 has a dot at all such points that are a distance of 1 unit from a point in Pattern 1. As the sequence continues, Pattern n + 1 has a dot at all such points that are a distance of 1 unit from a point in the previous pattern (Pattern n). This is true for all n ∈ ℕ.
(a)9 of the dots in Pattern 4 are shown on the co-ordinate diagram below. Draw in the missing dots to complete Pattern 4. Do not mark any points that are not in Pattern 4.
Add the missing dots of Pattern 4 directly on the grid
(b)In Pattern 2000, there are 4 points that are a distance of 2000 units from (0, 0). Write down the co-ordinates of these 4 points.
(b)
(c)What is the smallest value of n ∈ ℕ for which the point (4, 4) is in Pattern n? Do not draw on your diagram from part (a).
(c)
(d)Q(n) is the proportion of points with integer co-ordinates (a, b), where |a|, |b| ≤ n, that are in Pattern n of the sequence.
(i)Rearrange the formula t = 2n + 1 to write n in terms of t.
(d)(i)
Q(n) is given by the formula, for any n ∈ ℕ: Q(n) = (n + 1)2(2n + 1)2
(ii)By substituting your answer from (d)(i) into the expression for Q(n) above, show that Q(n) = t2 + 2t + 14t2.
(d)(ii)
(iii)Find the value of Q∞, which is given by the limit: Q∞ = limt→∞ t2 + 2t + 14t2.
(d)(iii)
(a)The first three terms of an arithmetic sequence are as follows, where p ∈ ℝ: T1 = 2p + 1
T2 = 5p − 3
T3 = 6p + 7 Find the value of p.
T2 = 5p − 3
T3 = 6p + 7 Find the value of p.
(a)
(b)G7 = 6 and G11 = 38 are the 7th and 11th terms of a geometric sequence, respectively. Find the two possible values of r, the common ratio of this sequence, where r ∈ ℝ.
(b)
(b)Find the value of the following limit, where n ∈ ℕ: limn→∞ [ nn + 1 + n + 1000n + (13)n ]
(b)
A triangle has a base of length 2 units and a perpendicular height of 8 units. The diagrams show T1, T2, and T3, the first three shapes in a sequence based on this triangle. For each n ∈ ℕ, the shape Tn is made up of n rectangles of equal height laid on top of each other; Tn is the collection of the smallest such rectangles that completely covers the triangle.
(a)Draw T4 in the grid (based on the triangle given on the grid).
(a)
(b)Show that the total area of the three rectangles in T3 is 323 square units.
(b)
(c)Find the total area of the n rectangles in Tn, for n ∈ ℕ. Give your answer in square units in terms of n, in its simplest form.
(c)
(d)The total area of the rectangles in the nth term of a different sequence is An = 8(n − 1)n, for n ∈ ℕ. Work out the first value of n for which An is greater than 95% of the area of the triangle (base 2, height 8).
(d)
(a)A sequence u1, u2, u3, ... is defined for n ∈ ℕ by: u1 = 2, u2 = 64, un+1 = √(unun−1) Write u3 in the form 2p, where p ∈ ℝ.
(a)
(b)The first three terms in an arithmetic sequence are as follows, where k ∈ ℝ: 5e−k, 13, 5ek
(i)By letting y = ek in this arithmetic sequence, show that 5y2 − 26y + 5 = 0.
(b)(i)
(ii)Use the equation in y to find the two possible values of k. Give each value in the form ln p or −ln p, where p ∈ ℕ.
(b)(ii)
Alex gets injections of a medicinal drug. Each injection has 15 mg. Each day the amount left in Alex's body from an injection decreases by 40%, so the amount (mg) left t days after a single injection is 15(0·6)t, where t ∈ ℝ.
(a)Find the amount of the drug left in Alex's body 2·5 days after a single 15 mg injection. Give your answer in mg, correct to 2 decimal places.
(a)
(b)How long after a single 15 mg injection will there be exactly 1 mg of the drug left in Alex's body? Give your answer in days, correct to 1 decimal place.
(b)
Alex is given a 15 mg injection at the same time every day for a long period.
(c)Explain why the total amount of the drug (mg) in Alex's body immediately after the 4th injection is 15 + 15(0·6) + 15(0·6)2 + 15(0·6)3.
(c)
(d)Find the total amount of the drug in Alex's body immediately after the 10th injection. Give your answer in mg, correct to 2 decimal places.
(d)
(e)Use the formula for the sum to infinity of a geometric series to estimate the amount of the drug (mg) in Alex's body after a long period of daily injections.
(e)
(f)Jessica also gets daily injections of d mg (d ∈ ℝ). Each day, the amount left from an injection decreases by 15%.
(i)Use the sum of a geometric series to show that the total amount (mg) in Jessica's body immediately after the nth injection (n ∈ ℕ) is 20d(1 − 0·85n)3.
(f)(i)
(ii)Immediately after the 7th injection, there are 50 mg of the drug in Jessica's body. Find the amount of the drug in one of Jessica's daily injections. Give your answer correct to the nearest mg.
(f)(ii)
(a)Prove using induction that 23n−1 + 3 is divisible by 7 for all n ∈ ℕ.
(a)
(b)p, p + 7, p + 14, p + 21, ... is an arithmetic sequence, where p ∈ ℕ.
(i)Find the nth term, Tn, in terms of n and p, where n ∈ ℕ.
(b)(i)
(ii)Find the smallest value of p for which 2021 is a term in the sequence.
(b)(ii)
The tip of the pendulum of a grandfather clock swings initially through an arc length of 45 cm. On each successive swing the length of the arc is 90% of the previous length.
(a)(i) Complete the table by filling in the missing lengths (Swing 1 = 45, Swing 3 = 72920).
| Swing | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Length of Arc (cm) | 45 | 729/20 |
(a)(i)
(ii)Tn = 45(0·9)n−1 is the arc length of swing n. Find the arc length of swing 25, correct to 1 decimal place.
(a)(ii)
(iii)Find the total distance travelled by the tip of the pendulum when it has completed swing 40. Give your answer in cm, correct to the nearest whole number.
(a)(iii)
(iv)Swing p is the first swing which has an arc length of less than 2 cm. Find the value of p.
(a)(iv)
(b)(i) If the length of the pendulum is 1 m, show that the angle θ of swing 1 is 26°, correct to the nearest degree.
(b)(i)
(ii)Hence, find the total accumulated angle that the pendulum swings through (the sum of all the angles it swings through until it stops). Give your answer correct to the nearest degree.
(b)(ii)
(iii)Hence, or otherwise, find the total distance travelled by the tip of the pendulum when it has moved through half of the total accumulated angle. Give your answer in cm, correct to the nearest integer.
(b)(iii)
(a)A number of the form 1 + 2 + 3 + … + n is sometimes called a triangular number because it can be represented as an equilateral triangle. The diagram shows the first three terms in the sequence of triangular numbers (T1 = 1, T2 = 3, T3 = 6).
(i)Complete the table to list the next five triangular numbers (T4 to T8).
(a)(i)
(ii)The nth triangular number can be found directly using the formula Tn = n(n + 1)2. Is 1275 a triangular number? Give a reason for your answer.
(a)(ii)
(b)(i)The (n + 1)th triangular number can be written as Tn+1 = Tn + (n + 1), where n ∈ ℕ. Write the expression n(n + 1)2 + (n + 1) as a single fraction in its simplest form.
(b)(i)
(ii)Prove that the sum of any two consecutive triangular numbers will always be a square number (a number in the form k2, where k ∈ ℕ).
(b)(ii)
(iii)Two consecutive triangular numbers sum to 12 544. Find the smaller of these two numbers.
(b)(iii)
(c)Some numbers are both triangular and square, for example 36. Leonhard Euler (1778) discovered the formula Nk = ((3 + 2√2)k − (3 − 2√2)k4√2)2, where Nk is the kth number that is both triangular and square. Use Euler's formula to find N3, the third number that is both triangular and square.
(c)
(d)Prove using induction that, for all n ∈ ℕ, the sum of the first n square numbers can be found using the formula 12 + 22 + 32 + … + n2 = n(n + 1)(2n + 1)6.
(d)
The closed line segment [0, 1] and the first three steps in the construction of the Cantor Set are shown. Step 1 removes the open middle third of [0, 1] leaving two closed line segments; Step 2 removes the middle third of the two remaining segments leaving four; Step 3 removes the middle third of the four remaining segments leaving eight. The process continues indefinitely. The set of points in [0, 1] that are not removed is the Cantor Set.
(a)(i)Complete the table to show the length of the line segment(s) removed at each step for the first 5 steps (Step 1 removes 13, Step 2 removes 29). Give your answers as fractions.
(a)(i)
(ii)Find the total length of all of the line segments removed from the initial line segment of length 1 unit, after a finite number (n) of steps. Give your answer in terms of n (a geometric series).
(a)(ii)
(iii)Find the total length removed, from the initial line segment, after an infinite number of steps.
(a)(iii)
(b)(i)Complete the table to identify the end-points labelled A to F in the diagram. Give your answers as fractions.
(b)(i)
(ii)Give a reason why 13 − 19 + 127 − 181 is a point in the Cantor Set.
(b)(ii)
(iii)The limit of the series 13 − 19 + 127 − … is a point in the Cantor Set. Find this point.
(b)(iii)
(a)The first three terms of a geometric series are x2, 5x − 8, and x + 8, where x ∈ ℝ. Use the common ratio to show that x3 − 17x2 + 80x − 64 = 0.
(a)
(b)If f(x) = x3 − 17x2 + 80x − 64, x ∈ ℝ, show that f(1) = 0, and find another value of x for which f(x) = 0.
(b)
(c)In the case of one of the values of x from part (b), the terms in part (a) will generate a geometric series with a finite sum to infinity. Find this value of x and hence find the sum to infinity.
(c)
(a)The Sieve of Sundaram is an infinite table of arithmetic sequences. The terms in the first 4 rows and first 4 columns are: row 1: 4, 7, 10, 13; row 2: 7, 12, 17, 22; row 3: 10, 17, 24, 31; row 4: 13, 22, 31, 40.
(i)Find the difference between the sums of the first 45 terms in the first two rows.
(a)(i)
(ii)Find the number which is in the 60th row and 70th column of the table.
(a)(ii)
(b)The first two terms of a sequence are a1 = 4 and a2 = 2. The general term is defined by an = an−1 − an−2, when n ≥ 3. Write out the next 6 terms of the sequence and hence find the value of a2019.
(b)
The diagram shows the first 4 steps of an infinite pattern which creates the Sierpinski Triangle. The sequence begins with a black equilateral triangle. Each step is formed by removing an equilateral triangle (joining the midpoints of the sides) from the centre of each black triangle in the previous step.
(a)Complete the table showing the number of black triangles at each of the first 4 steps and the fraction of the original triangle remaining (Step 0: 1 triangle, fraction 1; Step 2: fraction 916).
(a)
(b)(i)Write an expression in terms of n for the number of black triangles in step n of the pattern.
(b)(i)
(ii)Step k is the first step in which the number of black triangles exceeds one thousand million (1 × 109) for the first time. Find the value of k.
(b)(ii)
(c)(i)Step h is the first step in which the fraction of the original triangle remaining is less than 1100 of the original triangle. Find the value of h.
(c)(i)
(ii)What fraction of the original triangle remains after an infinite number of steps of the pattern?
(c)(ii)
(d)(i)The side length of the triangle in Step 0 is 1 unit. The total perimeter of all the black triangles in each step: Step 0 is 3, Step 2 is 274. Complete the table for the first 5 steps.
(d)(i)
(ii)Find the total perimeter of the black triangles in step 35 of the pattern. Give your answer correct to the nearest unit.
(d)(ii)
(iii)Use your answers to part (c)(ii) and part (d)(ii) to comment on the total area and the total perimeter of the black triangles in step n of the pattern, as n tends to infinity.
(d)(iii)
(a)The amount of a substance in a solution reduces exponentially over time. The percentage remaining is measured at the same time each day: Day 1: 95, Day 2: 42·75, Day 3: 19·2375, Day 4: 8·6569. Based on the data, estimate the first day on which the percentage will be less than 0·01% (a geometric sequence).
(a)
(b)A square has sides of length 2 cm. The midpoints of the sides are joined to form another square, and this process is continued. Find the sum to infinity of the perimeters of the squares. Give your answer in the form a + b√c cm, where a, b, c ∈ ℕ.
(b)
(a)Prove by induction that 8n − 1 is divisible by 7 for all n ∈ ℕ.
(a)
(a)At the first stage of a pattern, a point moves 4 units from the origin in the positive x-direction. It then turns left and moves 2 units parallel to the y-axis, then turns left and moves 1 unit parallel to the x-axis. At each stage after the first, the point turns left and moves half the distance of the previous stage (a geometric pattern).
(i)How many stages has the point completed when the total distance travelled along its path is 7·9375 units?
(a)(i)
(ii)Find the maximum distance the point can move, along its path, if it continues in this pattern indefinitely (sum to infinity).
(a)(ii)
(iii)Complete the table showing the changes to the x co-ordinate for the first nine moves. Hence find the x and y co-ordinates of the final position the point is approaching, if it continues indefinitely.
(a)(iii)
(b)A male bee comes from an unfertilised egg (a female parent, no male parent); a female bee comes from a fertilised egg (a female and a male parent).
(i)The diagram shows the ancestors of a male bee. His generation is G1; the diagram goes back to G4. Continue the diagram to G5.
(b)(i)
(ii)The number of ancestors in each generation is given by Gn+2 = Gn+1 + Gn, where G1 = 1 and G2 = 1 (the Fibonacci sequence). Calculate the number of ancestors in G6 and G7.
(b)(ii)
(iii)The number of ancestors can also be found using Gn = (1 + √5)n − (1 − √5)n2n√5. Use this formula to verify the number of ancestors in G3.
(b)(iii)
Mary threw a ball onto level ground from a height of 2 m. Each time the ball hit the ground it bounced back up to 34 of the height of the previous bounce (a geometric pattern).
(a)Complete the table to show the maximum height (in fraction form) reached by the ball on each of the first four bounces.
(a)
(b)Find, in metres, the total vertical distance (up and down) the ball had travelled when it hit the ground for the 5th time. Give your answer in fraction form.
(b)
(c)If the ball were to continue to bounce indefinitely, find, in metres, the total vertical distance it would travel (sum to infinity).
(c)
(a)Prove, by induction, that the sum of the first n natural numbers, 1 + 2 + 3 + … + n, is n(n + 1)2.
(a)
(b)Hence, or otherwise, prove that the sum of the first n even natural numbers, 2 + 4 + 6 + … + 2n, is n2 + n.
(b)
(c)Using the results from (a) and (b) above, find an expression for the sum of the first n odd natural numbers in its simplest form.
(c)
The nth term of a sequence is Tn = ln(an), where a > 0 and a is a constant.
(a)(i)Show that T1, T2 and T3 are in arithmetic sequence.
(a)(i)
(ii)Prove that the sequence is arithmetic and find the common difference.
(a)(ii)
(b)Find the value of a for which T1 + T2 + T3 + … + T98 + T99 + T100 = 10100.
(b)
(c)Verify that, for all values of a, T1 + T2 + T3 + … + T10 + 100d = T11 + T12 + T13 + … + T20, where d is the common difference of the sequence.
(c)
Questions reproduced from State Examinations Commission Leaving Certificate examination papers,
© State Examinations Commission (examinations.ie). Gathered here for study use.
© State Examinations Commission (examinations.ie). Gathered here for study use.