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← Problem-Solving Strategies levels

Problem-Solving Strategies / LEVEL 4 · DIFFICULTY 4/5

Combine Ideas

Recover totals, telescope logarithms and transform recurrences.

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

STAGE PS 4.1 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Recover the Totals

Useful preparation: Share an Altitude

Goal: Understand and apply recover the totals.

Before you begin: Share an Altitude

Understand the idea

A group average hides its total and size. Combining unequal groups requires recovering each total, adding them and dividing by the combined number of observations.

combined mean=n1 m1+n2 m2n1+n2\displaystyle \frac{n_{1}\, m_{1} + n_{2}\, m_{2}}{n_{1} + n_{2}}

Choose and carry out a method

Multiply each mean by its group size. Add those totals and divide by the sum of the group sizes.

Check the reasoning

The combined mean lies between the two original means and is pulled toward the larger group. Equal weighting is valid only for equal group sizes.

WORKED EXAMPLE 1

A group of 3 numbers has mean 6; another group of 6 numbers has mean 10. Find the mean of all the numbers together.

  1. Recover each total before combining groups of unequal size.
  2. Combined total=6⋅3+10⋅6\displaystyle 6 \cdot 3 + 10 \cdot 6; count=9.
  3. The mean is 263\displaystyle \frac{26}{3}. It lies between 6 and 10 and is closer to the larger group’s mean.

263\displaystyle \frac{26}{3}

WORKED EXAMPLE 2

A group of 4 numbers has mean 6; another group of 7 numbers has mean 10. Find the mean of all the numbers together.

  1. Recover each total before combining groups of unequal size.
  2. Combined total=6⋅4+10⋅7\displaystyle 6 \cdot 4 + 10 \cdot 7; count=11.
  3. The mean is 9411\displaystyle \frac{94}{11}. It lies between 6 and 10 and is closer to the larger group’s mean.

9411\displaystyle \frac{94}{11}

Common pitfalls

Possible mix-up: Average the two means directly.

Weight each mean by the number of observations it represents.

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

Explain why a larger group has more influence on the combined average.

Preview the eight practice prompts
  1. Problem 1

    A group of 6 numbers has mean 6; another group of 9 numbers has mean 10. Find the mean of all the numbers together.

    Practise this exact problem →
  2. Problem 2

    A group of 7 numbers has mean 6; another group of 10 numbers has mean 10. Find the mean of all the numbers together.

    Practise this exact problem →
  3. Problem 3

    A group of 8 numbers has mean 6; another group of 11 numbers has mean 10. Find the mean of all the numbers together.

    Practise this exact problem →
  4. Problem 4

    A group of 9 numbers has mean 6; another group of 12 numbers has mean 10. Find the mean of all the numbers together.

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  5. Problem 5

    A group of 10 numbers has mean 6; another group of 13 numbers has mean 10. Find the mean of all the numbers together.

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  6. Problem 6

    A group of 11 numbers has mean 6; another group of 14 numbers has mean 10. Find the mean of all the numbers together.

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  7. Problem 7New context

    12 containers each hold 6 liters and 15 each hold 10 liters. Find the average liters per container.

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  8. Problem 8New context

    13 containers each hold 6 liters and 16 each hold 10 liters. Find the average liters per container.

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Open stage PS 4.1 in the student workspace →

STAGE PS 4.2 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Telescope a Product

Useful preparation: Recover the Totals

Goal: Understand and apply telescope a product.

Before you begin: Recover the Totals

Understand the idea

Changing logarithm bases can turn complicated factors into adjacent exponent ratios. In a product of consecutive ratios, numerator factors cancel with denominators of neighboring factors.

∏ from k=1\displaystyle k = 1 to n (k+1)k=n+1\displaystyle \frac{n\, \left(k + 1\right)}{k} = n + 1

Choose and carry out a method

Rewrite log⁡2k (2k+1)\displaystyle \operatorname{log}_{2^{k}}\, \left(2^{k + 1}\right) as k+1k\displaystyle \frac{k + 1}{k}. Multiply the chain, retain its boundary factors and take the reciprocal if requested.

Check the reasoning

A reciprocal reverses the whole multiplier. Check the first and last indices carefully before deciding what survives.

WORKED EXAMPLE 1

Find the product log⁡2 (4)⋅log⁡4 (8)\displaystyle \operatorname{log}_{2}\, \left(4\right) \cdot \operatorname{log}_{4}\, \left(8\right)·…·log⁡24 (25)\displaystyle \operatorname{log}_{2^{4}}\, \left(2^{5}\right), then take its reciprocal. Enter the final reciprocal.

  1. Change each logarithm to base 2 and cancel neighboring factors.
  2. log⁡2k (2k+1)=k+1k\displaystyle \operatorname{log}_{2^{k}}\, \left(2^{k + 1}\right) = \frac{k + 1}{k}, so the product is 5.
  3. The reversing factor is 15=15\displaystyle \frac{1}{5} = \frac{1}{5}. A reciprocal undoes multiplication.

15\displaystyle \frac{1}{5}

WORKED EXAMPLE 2

Find the product log⁡2 (4)⋅log⁡4 (8)\displaystyle \operatorname{log}_{2}\, \left(4\right) \cdot \operatorname{log}_{4}\, \left(8\right)·…·log⁡25 (26)\displaystyle \operatorname{log}_{2^{5}}\, \left(2^{6}\right), then take its reciprocal. Enter the final reciprocal.

  1. Change each logarithm to base 2 and cancel neighboring factors.
  2. log⁡2k (2k+1)=k+1k\displaystyle \operatorname{log}_{2^{k}}\, \left(2^{k + 1}\right) = \frac{k + 1}{k}, so the product is 6.
  3. The reversing factor is 16=16\displaystyle \frac{1}{6} = \frac{1}{6}. A reciprocal undoes multiplication.

16\displaystyle \frac{1}{6}

Common pitfalls

Possible mix-up: A logarithm product equals a sum of exponents.

Convert each factor using change of base, then multiply.

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

What remains when the chain starts at k=3\displaystyle k = 3 instead of k=1\displaystyle k = 1?

Preview the eight practice prompts
  1. Problem 1

    Find the product log⁡2 (4)⋅log⁡4 (8)\displaystyle \operatorname{log}_{2}\, \left(4\right) \cdot \operatorname{log}_{4}\, \left(8\right)·…·log⁡27 (28)\displaystyle \operatorname{log}_{2^{7}}\, \left(2^{8}\right), then take its reciprocal. Enter the final reciprocal.

    Practise this exact problem →
  2. Problem 2

    Find the product log⁡2 (4)⋅log⁡4 (8)\displaystyle \operatorname{log}_{2}\, \left(4\right) \cdot \operatorname{log}_{4}\, \left(8\right)·…·log⁡28 (29)\displaystyle \operatorname{log}_{2^{8}}\, \left(2^{9}\right), then take its reciprocal. Enter the final reciprocal.

    Practise this exact problem →
  3. Problem 3

    Find the product log⁡2 (4)⋅log⁡4 (8)\displaystyle \operatorname{log}_{2}\, \left(4\right) \cdot \operatorname{log}_{4}\, \left(8\right)·…·log⁡29 (210)\displaystyle \operatorname{log}_{2^{9}}\, \left(2^{10}\right), then take its reciprocal. Enter the final reciprocal.

    Practise this exact problem →
  4. Problem 4

    Find the product log⁡2 (4)⋅log⁡4 (8)\displaystyle \operatorname{log}_{2}\, \left(4\right) \cdot \operatorname{log}_{4}\, \left(8\right)·…·log⁡210 (211)\displaystyle \operatorname{log}_{2^{10}}\, \left(2^{11}\right), then take its reciprocal. Enter the final reciprocal.

    Practise this exact problem →
  5. Problem 5

    Find the product log⁡2 (4)⋅log⁡4 (8)\displaystyle \operatorname{log}_{2}\, \left(4\right) \cdot \operatorname{log}_{4}\, \left(8\right)·…·log⁡211 (212)\displaystyle \operatorname{log}_{2^{11}}\, \left(2^{12}\right), then take its reciprocal. Enter the final reciprocal.

    Practise this exact problem →
  6. Problem 6

    Find the product log⁡2 (4)⋅log⁡4 (8)\displaystyle \operatorname{log}_{2}\, \left(4\right) \cdot \operatorname{log}_{4}\, \left(8\right)·…·log⁡212 (213)\displaystyle \operatorname{log}_{2^{12}}\, \left(2^{13}\right), then take its reciprocal. Enter the final reciprocal.

    Practise this exact problem →
  7. Problem 7New context

    A chain multiplies a quantity successively by 21\displaystyle \frac{2}{1}, 32\displaystyle \frac{3}{2}, …, 1413\displaystyle \frac{14}{13}. What factor reverses the entire chain?

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  8. Problem 8New context

    A chain multiplies a quantity successively by 21\displaystyle \frac{2}{1}, 32\displaystyle \frac{3}{2}, …, 1514\displaystyle \frac{15}{14}. What factor reverses the entire chain?

    Practise this exact problem →
Open stage PS 4.2 in the student workspace →

STAGE PS 4.3 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Transform a Recurrence

Useful preparation: Telescope a Product

Goal: Understand and apply transform a recurrence.

Before you begin: Telescope a Product

Understand the idea

Adding a fixed constant to every sequence term can turn an affine recurrence into a geometric one. The right shift absorbs the repeated added term.

an=(a0+1) 2n−1\displaystyle a_{n} = \left(a_{0} + 1\right)\, 2^{n} - 1

Choose and carry out a method

For an+1=2 an+1\displaystyle a_{n + 1} = 2\, a_{n} + 1, define bn=an+1\displaystyle b_{n} = a_{n} + 1. Then bn+1=2 bn\displaystyle b_{n + 1} = 2\, b_{n}; solve the geometric recurrence and subtract the shift at the end.

Check the reasoning

Check the original initial term and the first transition. Counting rounds from zero prevents an extra multiplication.

WORKED EXAMPLE 1

A sequence has a0=3\displaystyle a_{0} = 3 and an+1=2 an+1\displaystyle a_{n + 1} = 2\, a_{n} + 1. Find a4\displaystyle a_{4}.

  1. Shift by a constant to remove the added term.
  2. Let bn=an+1\displaystyle b_{n} = a_{n} + 1. Then bn+1=2 bn\displaystyle b_{n + 1} = 2\, b_{n} and b0=4\displaystyle b_{0} = 4.
  3. a4=(3+1)⋅24−1=63\displaystyle a_{4} = \left(3 + 1\right) \cdot 2^{4} - 1 = 63. Verify the first step before using the formula.

63

WORKED EXAMPLE 2

A sequence has a0=4\displaystyle a_{0} = 4 and an+1=2 an+1\displaystyle a_{n + 1} = 2\, a_{n} + 1. Find a5\displaystyle a_{5}.

  1. Shift by a constant to remove the added term.
  2. Let bn=an+1\displaystyle b_{n} = a_{n} + 1. Then bn+1=2 bn\displaystyle b_{n + 1} = 2\, b_{n} and b0=5\displaystyle b_{0} = 5.
  3. a5=(4+1)⋅25−1=159\displaystyle a_{5} = \left(4 + 1\right) \cdot 2^{5} - 1 = 159. Verify the first step before using the formula.

159

Common pitfalls

Possible mix-up: Shift only the initial term.

Transform the entire recurrence and undo the shift on the final value.

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

Find a shift that simplifies an+1=3 an+2\displaystyle a_{n + 1} = 3\, a_{n} + 2.

Preview the eight practice prompts
  1. Problem 1

    A sequence has a0=6\displaystyle a_{0} = 6 and an+1=2 an+1\displaystyle a_{n + 1} = 2\, a_{n} + 1. Find a3\displaystyle a_{3}.

    Practise this exact problem →
  2. Problem 2

    A sequence has a0=7\displaystyle a_{0} = 7 and an+1=2 an+1\displaystyle a_{n + 1} = 2\, a_{n} + 1. Find a4\displaystyle a_{4}.

    Practise this exact problem →
  3. Problem 3

    A sequence has a0=8\displaystyle a_{0} = 8 and an+1=2 an+1\displaystyle a_{n + 1} = 2\, a_{n} + 1. Find a5\displaystyle a_{5}.

    Practise this exact problem →
  4. Problem 4

    A sequence has a0=9\displaystyle a_{0} = 9 and an+1=2 an+1\displaystyle a_{n + 1} = 2\, a_{n} + 1. Find a6\displaystyle a_{6}.

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  5. Problem 5

    A sequence has a0=10\displaystyle a_{0} = 10 and an+1=2 an+1\displaystyle a_{n + 1} = 2\, a_{n} + 1. Find a3\displaystyle a_{3}.

    Practise this exact problem →
  6. Problem 6

    A sequence has a0=11\displaystyle a_{0} = 11 and an+1=2 an+1\displaystyle a_{n + 1} = 2\, a_{n} + 1. Find a4\displaystyle a_{4}.

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  7. Problem 7New context

    A token pile starts with 12 tokens. Each round doubles the pile and adds one token. How many tokens are present after 5 rounds?

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  8. Problem 8New context

    A token pile starts with 13 tokens. Each round doubles the pile and adds one token. How many tokens are present after 6 rounds?

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Open stage PS 4.3 in the student workspace →
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