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Problem-Solving Strategies / LEVEL 2 · DIFFICULTY 2/5

Exploit Structure

Balance extremes, cancel terms and justify patterns.

3 stages · 24 practice problems · two 6-question assessment forms

Choose an island to read its lesson.

  1. MINI QUEST PS 2.1Balance an ExtremeRead the lesson
  2. MINI QUEST PS 2.2Telescoping SumsRead the lesson
  3. MINI QUEST PS 2.3Patterns and InductionRead the lesson
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STAGE PS 2.1 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Balance an Extreme

Useful preparation: Force a Repetition

Goal: Understand and apply balance an extreme.

Before you begin: Force a Repetition

Understand the idea

With a fixed sum, the sum of two squares measures how unequal the two numbers are. Moving one unit from the larger integer to the smaller decreases the total whenever their gap exceeds one.

fixed sum ⇒ minimize |x−y| to minimize x²+y²

Choose and carry out a method

Repeatedly balance the pair until it differs by at most one. For an odd total, use the two neighboring integers around half the total.

Check the reasoning

Verify that both values satisfy the integer and nonnegativity conditions. The identity x²+y²=((x+y)²+(x−y)²)/2 explains the minimum.

WORKED EXAMPLE 1

Two nonnegative integers have sum 11. Find the smallest possible sum of their squares.

  1. Move one unit from the larger number to the smaller while their gap exceeds one.
  2. The closest integer pair is 5 and 6, so the sum of squares is 5²+6².
  3. The minimum is 61. A more uneven split increases the squared difference.

61

WORKED EXAMPLE 2

Two nonnegative integers have sum 13. Find the smallest possible sum of their squares.

  1. Move one unit from the larger number to the smaller while their gap exceeds one.
  2. The closest integer pair is 6 and 7, so the sum of squares is 6²+7².
  3. The minimum is 85. A more uneven split increases the squared difference.

85

Common pitfalls

Possible mix-up: Put everything in one variable for the minimum.

That maximizes imbalance; the minimum is attained by the closest pair.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Calculate the change in squared sum when one unit moves from the larger value to the smaller.

Preview the eight practice prompts
  1. Two nonnegative integers have sum 17. Find the smallest possible sum of their squares.
  2. Two nonnegative integers have sum 19. Find the smallest possible sum of their squares.
  3. Two nonnegative integers have sum 21. Find the smallest possible sum of their squares.
  4. Two nonnegative integers have sum 23. Find the smallest possible sum of their squares.
  5. Two nonnegative integers have sum 25. Find the smallest possible sum of their squares.
  6. Two nonnegative integers have sum 27. Find the smallest possible sum of their squares.
  7. 29 tasks are split between two teams. The workload penalty is the square of each team’s task count, added together. Find the smallest penalty. New context
  8. 31 tasks are split between two teams. The workload penalty is the square of each team’s task count, added together. Find the smallest penalty. New context
Open stage PS 2.1 in the student workspace →

STAGE PS 2.2 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Telescoping Sums

Useful preparation: Balance an Extreme

Goal: Understand and apply telescoping sums.

Before you begin: Balance an Extreme

Understand the idea

A fraction can sometimes be split into a difference whose neighboring terms cancel in a sum. The main work then lies in the first and last terms rather than the long middle.

Σ from k=1 to n 1/[k(k+1)]=1−1/(n+1)

Choose and carry out a method

Rewrite 1/[k(k+1)] as 1/k−1/(k+1). Write the first few and last few terms before canceling.

Check the reasoning

Do not cancel across addition without an explicit opposite term. Retain both endpoints and check the formula on a one-term sum.

WORKED EXAMPLE 1

Find the exact sum 1/(1·2)+1/(2·3)+…+1/(4·5).

  1. Split each fraction into a difference of consecutive reciprocals.
  2. 1/[k(k+1)]=1/k-1/(k+1); all interior terms cancel.
  3. The sum is 1-1/5=4/5. Keep both boundary terms.

4/5

WORKED EXAMPLE 2

Find the exact sum 1/(1·2)+1/(2·3)+…+1/(5·6).

  1. Split each fraction into a difference of consecutive reciprocals.
  2. 1/[k(k+1)]=1/k-1/(k+1); all interior terms cancel.
  3. The sum is 1-1/6=5/6. Keep both boundary terms.

5/6

Common pitfalls

Possible mix-up: Every term cancels, giving zero.

The first positive and last negative boundary terms remain.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Derive the fraction decomposition by putting the difference over a common denominator.

Preview the eight practice prompts
  1. Find the exact sum 1/(1·2)+1/(2·3)+…+1/(7·8).
  2. Find the exact sum 1/(1·2)+1/(2·3)+…+1/(8·9).
  3. Find the exact sum 1/(1·2)+1/(2·3)+…+1/(9·10).
  4. Find the exact sum 1/(1·2)+1/(2·3)+…+1/(10·11).
  5. Find the exact sum 1/(1·2)+1/(2·3)+…+1/(11·12).
  6. Find the exact sum 1/(1·2)+1/(2·3)+…+1/(12·13).
  7. A process adds 1/[k(k+1)] liters on step k, for k=1 through 13. Find its total addition in liters. New context
  8. A process adds 1/[k(k+1)] liters on step k, for k=1 through 14. Find its total addition in liters. New context
Open stage PS 2.2 in the student workspace →

STAGE PS 2.3 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Patterns and Induction

Useful preparation: Telescoping Sums

Goal: Understand and apply patterns and induction.

Before you begin: Telescoping Sums

Understand the idea

A pattern suggested by small cases needs a reason that extends to every case. Consecutive square areas differ by consecutive odd numbers, linking a dot picture with an algebraic induction step.

1+3+…+(2n−1)=n²

Choose and carry out a method

Verify the first case. Assuming the first n odd numbers sum to n², add the next odd number 2n+1 to obtain (n+1)².

Check the reasoning

The induction step must use the next term correctly. Testing many examples alone does not establish all positive integers.

WORKED EXAMPLE 1

Find the sum of the first 6 positive odd integers.

  1. Adding the next odd number extends a square to the next larger square.
  2. Sₙ=n²; the induction step is n²+(2n+1)=(n+1)².
  3. S_6=36. The nth odd number is 2n-1, not n.

36

WORKED EXAMPLE 2

Find the sum of the first 7 positive odd integers.

  1. Adding the next odd number extends a square to the next larger square.
  2. Sₙ=n²; the induction step is n²+(2n+1)=(n+1)².
  3. S_7=49. The nth odd number is 2n-1, not n.

49

Common pitfalls

Possible mix-up: Several successful examples prove the identity.

Supply a base case and a step covering every transition.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Draw the L-shaped layer that turns an n-by-n square into an (n+1)-by-(n+1) square.

Preview the eight practice prompts
  1. Find the sum of the first 9 positive odd integers.
  2. Find the sum of the first 10 positive odd integers.
  3. Find the sum of the first 11 positive odd integers.
  4. Find the sum of the first 12 positive odd integers.
  5. Find the sum of the first 13 positive odd integers.
  6. Find the sum of the first 14 positive odd integers.
  7. A square dot picture grows by adding the next odd number of dots in each layer, starting with one dot. How many dots are present after 15 layers? New context
  8. A square dot picture grows by adding the next odd number of dots in each layer, starting with one dot. How many dots are present after 16 layers? New context
Open stage PS 2.3 in the student workspace →
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