Identities involving powers and integers | Year 12

01 / LEARN

The idea, step by step

Track the exponent

Use a^(k+1)=a times a^k to connect consecutive powers. For alternating signs, (-1)^(k+1) is the negative of (-1)^k. A lost minus sign can invalidate a divisibility argument.

A geometric sum starting with exponent zero has final exponent n-1 when it has n terms. The next term after k terms is a^k.

Match the operation

Extend a product by multiplication and a sum by addition. In a weighted power sum, combine terms carrying the same power before matching the target. The logical proof structure stays unchanged.

These are explicit identities, not induction on an assumed recursive sequence. Any cancellation must involve a nonzero factor, which follows here from the positive-integer domain.

A common trap

Adding a factor to extend a product.

Instead: Multiply by the new factor; addition extends a sum.

02 / WORKED EXAMPLES

See the thinking

Example 1

A geometric sum

Prove $1 + 3 + 3^{2} + \dots + 3^{n - 1} = (3^{n} - 1) / 2$ for $n \geq 1$ .

$n = 1$ gives $(3 - 1) / 2 = 1$ .
Assume the sum to $k$ is $(3^{k} - 1) / 2$ .
Add $3^{k}$ .
The common numerator is $3^{k} - 1 + 2 (3^{k}) = 3^{k + 1} - 1$ .
Conclude all positive integers by induction.

1 + 3 + 3^{2} + \dots + 3^{n - 1} = (3^{n} - 1) / 2

is true throughout the stated domain.

Example 2

A weighted power identity

Prove $\sum_{r = 1}^{n} r 2^{r - 1} = (n - 1) 2^{n} + 1$ for $n \geq 1$ .

At $n = 1$ , both sides are $1$ .
Assume the sum to $k$ is $(k - 1) 2^{k} + 1$ .
Add $(k + 1) 2^{k}$ .
The power terms combine to $2 k 2^{k} = k 2^{k + 1}$ , giving the required formula.
Conclude all positive integers by induction.

\sum_{r = 1}^{n} r 2^{r - 1} = (n - 1) 2^{n} + 1

is true throughout the stated domain.

Check your understanding

What follows the first k terms of $1 + 3 + \dots + 3^{n - 1}$ ?

Show answer

$3^{k}$ .

The first exponent is zero.

03 / WORKSHEET

Your turn

6 questions · 36 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Prove $1 + 2 + 2^{2} + \dots + 2^{n - 1} = (2^{n} - 1)$ for all integers $n \geq 1$ .
6 marks
1. (a)
  Verify the first case and state the induction hypothesis.
  2 marks
2. (b)
  Prove the next case and conclude.
  4 marks
Prove $1 + 3 + 3^{2} + \dots + 3^{n - 1} = (3^{n} - 1) / 2$ for all integers $n \geq 1$ .
6 marks
1. (a)
  Verify the first case and state the induction hypothesis.
  2 marks
2. (b)
  Prove the next case and conclude.
  4 marks
Prove $\sum_{r = 1}^{n} r 2^{r - 1} = (n - 1) 2^{n} + 1$ for all integers $n \geq 1$ .
6 marks
1. (a)
  Verify the first case and state the induction hypothesis.
  2 marks
2. (b)
  Prove the next case and conclude.
  4 marks
Prove $\prod_{r = 1}^{n} (r + 1) / r = n + 1$ for all integers $n \geq 1$ .
6 marks
1. (a)
  Verify the first case and state the induction hypothesis.
  2 marks
2. (b)
  Prove the next case and conclude.
  4 marks
Prove $\prod_{r = 1}^{n} (2 r - 1) / (2 r + 1) = 1 / (2 n + 1)$ for all integers $n \geq 1$ .
6 marks
1. (a)
  Verify the first case and state the induction hypothesis.
  2 marks
2. (b)
  Prove the next case and conclude.
  4 marks
Prove $\sum_{r = 1}^{n} {(2 r - 1)}^{2} = n (4 n^{2} - 1) / 3$ for all integers $n \geq 1$ .
6 marks
1. (a)
  Verify the first case and state the induction hypothesis.
  2 marks
2. (b)
  Prove the next case and conclude.
  4 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Prove $1 + 2 + 2^{2} + \dots + 2^{n - 1} = (2^{n} - 1)$ for all integers $n \geq 1$ .
6 marks
(a) $n = 1$ gives $1 = 1$ .
(a) Assume the sum to $k$ is $(2^{k} - 1)$ .
(b) Add $2^{k}$ .
(b) The new total is $2^{k} - 1 + 2^{k} = 2^{k + 1} - 1$ .
(b) The true first case and successor implication establish the full domain by mathematical induction.
1. (a)
  Verify the first case and state the induction hypothesis.
  2 marks
  $n = 1$ gives $1 = 1$ .
  Assume the sum to $k$ is $(2^{k} - 1)$ .
  $n = 1$ gives $1 = 1$ . Assume the sum to $k$ is $(2^{k} - 1)$ .
2. (b)
  Prove the next case and conclude.
  4 marks
  Add $2^{k}$ .
  The new total is $2^{k} - 1 + 2^{k} = 2^{k + 1} - 1$ .
  The true first case and successor implication establish the full domain by mathematical induction.
  The statement holds for all positive integers.
(a) $n = 1$ gives $1 = 1$ . Assume the sum to $k$ is $(2^{k} - 1)$ . (b) The statement holds for all positive integers.
Prove $1 + 3 + 3^{2} + \dots + 3^{n - 1} = (3^{n} - 1) / 2$ for all integers $n \geq 1$ .
6 marks
(a) $n = 1$ gives $(3 - 1) / 2 = 1$ .
(a) Assume the sum to $k$ is $(3^{k} - 1) / 2$ .
(b) Add $3^{k}$ .
(b) The common numerator is $3^{k} - 1 + 2 (3^{k}) = 3^{k + 1} - 1$ .
(b) The true first case and successor implication establish the full domain by mathematical induction.
1. (a)
  Verify the first case and state the induction hypothesis.
  2 marks
  $n = 1$ gives $(3 - 1) / 2 = 1$ .
  Assume the sum to $k$ is $(3^{k} - 1) / 2$ .
  $n = 1$ gives $(3 - 1) / 2 = 1$ . Assume the sum to $k$ is $(3^{k} - 1) / 2$ .
2. (b)
  Prove the next case and conclude.
  4 marks
  Add $3^{k}$ .
  The common numerator is $3^{k} - 1 + 2 (3^{k}) = 3^{k + 1} - 1$ .
  The true first case and successor implication establish the full domain by mathematical induction.
  The statement holds for all positive integers.
(a) $n = 1$ gives $(3 - 1) / 2 = 1$ . Assume the sum to $k$ is $(3^{k} - 1) / 2$ . (b) The statement holds for all positive integers.
Prove $\sum_{r = 1}^{n} r 2^{r - 1} = (n - 1) 2^{n} + 1$ for all integers $n \geq 1$ .
6 marks
(a) At $n = 1$ , both sides are $1$ .
(a) Assume the sum to $k$ is $(k - 1) 2^{k} + 1$ .
(b) Add $(k + 1) 2^{k}$ .
(b) The power terms combine to $2 k 2^{k} = k 2^{k + 1}$ , giving the required formula.
(b) The true first case and successor implication establish the full domain by mathematical induction.
1. (a)
  Verify the first case and state the induction hypothesis.
  2 marks
  At $n = 1$ , both sides are $1$ .
  Assume the sum to $k$ is $(k - 1) 2^{k} + 1$ .
  At $n = 1$ , both sides are $1$ . Assume the sum to $k$ is $(k - 1) 2^{k} + 1$ .
2. (b)
  Prove the next case and conclude.
  4 marks
  Add $(k + 1) 2^{k}$ .
  The power terms combine to $2 k 2^{k} = k 2^{k + 1}$ , giving the required formula.
  The true first case and successor implication establish the full domain by mathematical induction.
  The statement holds for all positive integers.
(a) At $n = 1$ , both sides are $1$ . Assume the sum to $k$ is $(k - 1) 2^{k} + 1$ . (b) The statement holds for all positive integers.
Prove $\prod_{r = 1}^{n} (r + 1) / r = n + 1$ for all integers $n \geq 1$ .
6 marks
(a) At $n = 1$ , the product is $2$ .
(a) Assume the product to $k$ is $k + 1$ .
(b) Multiply by $(k + 2) / (k + 1)$ .
(b) The product becomes $k + 2$ .
(b) The true first case and successor implication establish the full domain by mathematical induction.
1. (a)
  Verify the first case and state the induction hypothesis.
  2 marks
  At $n = 1$ , the product is $2$ .
  Assume the product to $k$ is $k + 1$ .
  At $n = 1$ , the product is $2$ . Assume the product to $k$ is $k + 1$ .
2. (b)
  Prove the next case and conclude.
  4 marks
  Multiply by $(k + 2) / (k + 1)$ .
  The product becomes $k + 2$ .
  The true first case and successor implication establish the full domain by mathematical induction.
  The statement holds for all positive integers.
(a) At $n = 1$ , the product is $2$ . Assume the product to $k$ is $k + 1$ . (b) The statement holds for all positive integers.
Prove $\prod_{r = 1}^{n} (2 r - 1) / (2 r + 1) = 1 / (2 n + 1)$ for all integers $n \geq 1$ .
6 marks
(a) At $n = 1$ , the product is $1 / 3$ .
(a) Assume the product to $k$ is $1 / (2 k + 1)$ .
(b) Multiply by $(2 k + 1) / (2 k + 3)$ .
(b) The result is $1 / (2 k + 3) = 1 / [2 (k + 1) + 1]$ .
(b) The true first case and successor implication establish the full domain by mathematical induction.
1. (a)
  Verify the first case and state the induction hypothesis.
  2 marks
  At $n = 1$ , the product is $1 / 3$ .
  Assume the product to $k$ is $1 / (2 k + 1)$ .
  At $n = 1$ , the product is $1 / 3$ . Assume the product to $k$ is $1 / (2 k + 1)$ .
2. (b)
  Prove the next case and conclude.
  4 marks
  Multiply by $(2 k + 1) / (2 k + 3)$ .
  The result is $1 / (2 k + 3) = 1 / [2 (k + 1) + 1]$ .
  The true first case and successor implication establish the full domain by mathematical induction.
  The statement holds for all positive integers.
(a) At $n = 1$ , the product is $1 / 3$ . Assume the product to $k$ is $1 / (2 k + 1)$ . (b) The statement holds for all positive integers.
Prove $\sum_{r = 1}^{n} {(2 r - 1)}^{2} = n (4 n^{2} - 1) / 3$ for all integers $n \geq 1$ .
6 marks
(a) At $n = 1$ , both sides equal $1$ .
(a) Assume the formula holds at $n = k$ .
(b) Add ${(2 k + 1)}^{2}$ .
(b) The numerator becomes $4 k^{3} + 12 k^{2} + 11 k + 3 = (k + 1) (4 k^{2} + 8 k + 3)$ .
(b) This equals $(k + 1) [4 {(k + 1)}^{2} - 1]$ , as required.
(b) The true first case and successor implication establish the full domain by mathematical induction.
1. (a)
  Verify the first case and state the induction hypothesis.
  2 marks
  At $n = 1$ , both sides equal $1$ .
  Assume the formula holds at $n = k$ .
  At $n = 1$ , both sides equal $1$ . Assume the formula holds at $n = k$ .
2. (b)
  Prove the next case and conclude.
  4 marks
  Add ${(2 k + 1)}^{2}$ .
  The numerator becomes $4 k^{3} + 12 k^{2} + 11 k + 3 = (k + 1) (4 k^{2} + 8 k + 3)$ .
  This equals $(k + 1) [4 {(k + 1)}^{2} - 1]$ , as required.
  The true first case and successor implication establish the full domain by mathematical induction.
  The statement holds for all positive integers.
(a) At $n = 1$ , both sides equal $1$ . Assume the formula holds at $n = k$ . (b) The statement holds for all positive integers.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist mathematical induction. Explicit summation, power and divisibility identities; induction of inequalities or recursive formulae is not required.

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