Skip to content

Special Sequences: all problems

Focus mode: every practice problem from the chapter, one after another.

0 / 17 solved

Problem 17.2.1

AMC 8

What is the value of the product

(1⋅32⋅2)(2⋅43⋅3)(3⋅54⋅4)⋯(97⋅9998⋅98)(98⋅10099⋅99)?\left(\frac{1 \cdot 3}{2 \cdot 2}\right)\left(\frac{2 \cdot 4}{3 \cdot 3}\right)\left(\frac{3 \cdot 5}{4 \cdot 4}\right) \cdots \left(\frac{97 \cdot 99}{98 \cdot 98}\right)\left(\frac{98 \cdot 100}{99 \cdot 99}\right)?

Product of 5 fractions for k = 1 to 5: after cancelling, 7 / 2 · 6 = 7/12

last k5
1·32·2
×2·43·3
×3·54·4
×4·65·5
×5·76·6
0 of 8 cancellations

0 of 8 cancellations done

Problem 17.2.2

AMC 10

Let f(x)=x2(1−x)2f(x) = x^2(1 - x)^2. What is the value of the sum

f(12019)−f(22019)+f(32019)−f(42019)+⋯+f(20172019)−f(20182019)?f\left(\frac{1}{2019}\right) - f\left(\frac{2}{2019}\right) + f\left(\frac{3}{2019}\right) - f\left(\frac{4}{2019}\right) + \cdots + f\left(\frac{2017}{2019}\right) - f\left(\frac{2018}{2019}\right)?

Problem 17.2.3

AMC 10

Seven cubes, whose volumes are 11, 88, 2727, 6464, 125125, 216216, and 343343 cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?

1² + 2² + 3² + 4² + 5² + 6² + 7² = 140; 3 copies fill a 28 × 15 rectangle

n7
copies3

Three copies of the sum fill a 28 × 15 rectangle, so the sum is a third of it.

1² + 2² + 3² + 4² + 5² + 6² + 7² = 140
12+22+⋯+n2=n(n+1)(2n+1)6  ⇒  7⋅8⋅156=1401^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6} \;\Rightarrow\; \frac{7 \cdot 8 \cdot 15}{6} = 140

1² + 2² + 3² + 4² + 5² + 6² + 7² = 140; 3 copies fill a 28 × 15 rectangle

Problem 17.2.4

AMC 12

Rows 11, 22, 33, 44, and 55 of a triangular array of integers are shown below:

1111311551171171\begin{array}{ccccccccc} &&&& 1 &&&& \\ &&& 1 && 1 &&& \\ && 1 && 3 && 1 && \\ & 1 && 5 && 5 && 1 & \\ 1 && 7 && 11 && 7 && 1 \end{array}

Each row after the first row is formed by placing a 11 at each end of the row, and each interior entry is 11 greater than the sum of the two numbers diagonally above it in the previous row. What is the units digit of the sum of the 20232023 numbers in the 20232023rd row?

Problem 17.2.5

AMC 10Challenge

Hiram's algebra notes are 5050 pages long and are printed on 2525 sheets of paper; the first sheet contains pages 11 and 22, the second sheet contains pages 33 and 44, and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly 1919. How many sheets were borrowed?

Problem 17.2.6(additional)

AHSME

If Tn=1+2+3+⋯+nT_n = 1 + 2 + 3 + \cdots + n and

Pn=T2T2−1⋅T3T3−1⋅T4T4−1⋅⋯⋅TnTn−1P_n = \frac{T_2}{T_2 - 1} \cdot \frac{T_3}{T_3 - 1} \cdot \frac{T_4}{T_4 - 1} \cdot \cdots \cdot \frac{T_n}{T_n - 1}

for n=2,3,4,…n = 2, 3, 4, \ldots, then P1991P_{1991} is closest to which of the following numbers?

Product of 5 fractions for k = 2 to 6: after cancelling, 3 · 6 / 8 = 9/4

last k6
2·31·4
×3·42·5
×4·53·6
×5·64·7
×6·75·8
0 of 8 cancellations

0 of 8 cancellations done

Click a choice or type its letter (A–E).

Problem 17.2.7(additional)

AMC 12

Let AA be the set of positive integers that have no prime factors other than 22, 33, or 55. The infinite sum

11+12+13+14+15+16+18+19+110+112+115+116+118+120+⋯\frac11 + \frac12 + \frac13 + \frac14 + \frac15 + \frac16 + \frac18 + \frac19 + \frac1{10} + \frac1{12} + \frac1{15} + \frac1{16} + \frac1{18} + \frac1{20} + \cdots

of the reciprocals of the elements of AA can be expressed as mn\frac{m}{n}, where mm and nn are relatively prime positive integers. What is m+nm + n?

Problem 17.2.8(additional)

AHSME

Find the sum 11(3)+13(5)+⋯+1(2n−1)(2n+1)+⋯+1255(257)\dfrac{1}{1(3)} + \dfrac{1}{3(5)} + \cdots + \dfrac{1}{(2n-1)(2n+1)} + \cdots + \dfrac{1}{255(257)}.

Sum of 1/((2k − 1)(2k + 1)) for k = 1 to 4: each term is 1/2 × (1/(2k − 1) − 1/(2k + 1)), and the middle cancels, leaving 1/2 × (1/1 − 1/9) = 4/9

last k4
1/(1·3)12(11 − 13)
+1/(3·5)12(13 − 15)
+1/(5·7)12(15 − 17)
+1/(7·9)12(17 − 19)
0 of 3 cancellations

0 of 3 cancellations done

Problem 17.2.9(additional)

AMC 12B

Consider a decreasing sequence of nn positive integers x1>x2>⋯>xnx_1 > x_2 > \cdots > x_n that satisfies the following conditions: The average of the first 33 terms in the sequence is 20252025. For all 4<k<n4 < k < n, the average of the first kk terms is 11 less than the average of the first k−1k - 1 terms. What is the greatest possible value of nn?

Problem 17.2.10(additional)

AMC 10

The Fibonacci sequence 1,1,2,3,5,8,13,21,…1, 1, 2, 3, 5, 8, 13, 21, \ldots starts with two 11s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?

Problem 17.2.11(additional)

AIME

Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.

Problem 17.2.12(additional)

Evaluate

(1−(13)2)(1−(14)2)(1−(15)2)⋯\left(1 - \left(\frac13\right)^2\right)\left(1 - \left(\frac14\right)^2\right)\left(1 - \left(\frac15\right)^2\right) \cdots

Product of 5 fractions for k = 3 to 7: after cancelling, 2 · 8 / 3 · 7 = 16/21

last k7
2·43·3
×3·54·4
×4·65·5
×5·76·6
×6·87·7
0 of 8 cancellations

0 of 8 cancellations done

Problem 17.2.13(additional)

AIME

A sequence of integers a1,a2,a3,…a_1, a_2, a_3, \ldots is chosen so that an=an−1−an−2a_n = a_{n-1} - a_{n-2} for each n≥3n \ge 3. What is the sum of the first 20012001 terms of this sequence if the sum of the first 14921492 terms is 19851985, and the sum of the first 19851985 terms is 14921492?

a_n = a_(n−1) − a_(n−2): the terms repeat 2, 5, 3, -2, -5, -3 (length 6)

a12
a25

an=an−1−an−2a_n = a_{n-1} - a_{n-2}

  1. a1a_{1}2S = 2
  2. a2a_{2}5S = 7
  3. a3a_{3}3S = 10
  4. a4a_{4}-2S = 8
  5. a5a_{5}-5S = 3
  6. a6a_{6}-3S = 0
  7. a7a_{7}2S = 2
  8. a8a_{8}5S = 7
cycle length 6

a_n = a_(n−1) − a_(n−2): the terms repeat 2, 5, 3, -2, -5, -3 (length 6)

Problem 17.2.14(additional)

AMC 12

Set u0=14u_0 = \frac14, and for k≥0k \ge 0 let uk+1u_{k+1} be determined by the recurrence

uk+1=2uk−2uk2.u_{k+1} = 2u_k - 2u_k^2.

This sequence tends to a limit; call it LL. What is the least value of kk such that

∣uk−L∣≤121000?|u_k - L| \le \frac{1}{2^{1000}}?

Problem 17.2.15(additional)

AMC 10Challenge

The sum of the first mm positive odd integers is 212212 more than the sum of the first nn positive even integers. What is the sum of all possible values of nn?

Problem 17.2.16(additional)

AIMEChallenge

The sequence {an}\{a_n\} is defined by

a0=1,a1=1,andan=an−1+an−12an−2 for n≥2.a_0 = 1, \quad a_1 = 1, \quad \text{and} \quad a_n = a_{n-1} + \frac{a_{n-1}^2}{a_{n-2}} \text{ for } n \ge 2.

The sequence {bn}\{b_n\} is defined by

b0=1,b1=3,andbn=bn−1+bn−12bn−2 for n≥2.b_0 = 1, \quad b_1 = 3, \quad \text{and} \quad b_n = b_{n-1} + \frac{b_{n-1}^2}{b_{n-2}} \text{ for } n \ge 2.

Find b32a32\dfrac{b_{32}}{a_{32}}.

Problem 17.2.17(additional)

AIMEChallenge

The terms of the sequence (ai)(a_i) defined by an+2=an+20091+an+1a_{n+2} = \dfrac{a_n + 2009}{1 + a_{n+1}} for n≥1n \ge 1 are positive integers. Find the minimum possible value of a1+a2a_1 + a_2.