Identities involving powers and integers | Year 12

01 / LEARN

Understand the idea.

Track the exponent

Count terms separately from exponents. The $n$ terms $1, a, \dots, a^{n - 1}$ begin at exponent zero, so extending the first $k$ terms adds $a^{k}$ , not $a^{k + 1}$ .

In a weighted term such as $r 2^{r - 1}$ , substitute the new index in both places: the next term is $(k + 1) 2^{k}$ . Keep its coefficient attached to the power while simplifying.

Match the operation

Use $a^{k + 1} = a a^{k}$ to combine terms with a common power. Factoring $2^{k}$ , for example, lets you add the coefficients before converting back to a successor power.

The algebra sits inside the same induction structure: verify the first case, assume the identity at arbitrary $k$ , add the new term, and reach the formula at $k + 1$ . State the final domain explicitly.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A geometric sum

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

Worked solution

1
Verify the one-term sum.
$n = 1 : 1 = \frac{3^{1} - 1}{2} .$
The final exponent is zero when there is one term.
2
Assume the first $k$ terms.
$\sum_{r = 0}^{k - 1} 3^{r} = \frac{3^{k} - 1}{2}, k \in ℤ, k \geq 1.$
The upper exponent is one less than the number of terms.
3
Append the next power.
$\sum_{r = 0}^{k} 3^{r} = \sum_{r = 0}^{k - 1} 3^{r} + 3^{k} = \frac{3^{k} - 1}{2} + 3^{k} .$
The original $k$ terms are exactly the sum covered by the hypothesis.
4
Combine the common power.
$\begin{aligned} \sum_{r = 0}^{k} 3^{r} & = \frac{3^{k} - 1 + 2 \cdot 3^{k}}{2} \\ = \frac{3 \cdot 3^{k} - 1}{2} \\ = \frac{3^{k + 1} - 1}{2} . \end{aligned}$
Three copies of $3^{k}$ equal $3^{k + 1}$ .
5
Conclude the identity.
$\sum_{r = 0}^{n - 1} 3^{r} = \frac{3^{n} - 1}{2} (n \in ℤ, n \geq 1) .$
The base and successor implication prove the required statement by induction.

Answer

$1 + 3 + 3^{2} + \dots + 3^{n - 1} = (3^{n} - 1) / 2$ is true throughout the stated domain.

EXAMPLE 02

A weighted power identity

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

Worked solution

1
Check the first weighted term.
$n = 1 : 1 \cdot 2^{0} = 1 = (1 - 1) 2^{1} + 1.$
Both sides equal one.
2
Assume the identity at $k$ .
$\sum_{r = 1}^{k} r 2^{r - 1} = (k - 1) 2^{k} + 1, k \in ℤ, k \geq 1.$
The complete right side, including its constant term, belongs to the hypothesis.
3
Add the new weighted term.
$\begin{aligned} \sum_{r = 1}^{k + 1} r 2^{r - 1} & = \sum_{r = 1}^{k} r 2^{r - 1} + (k + 1) 2^{k} \\ = (k - 1) 2^{k} + 1 + (k + 1) 2^{k} . \end{aligned}$
At $r = k + 1$ , the coefficient becomes $k + 1$ and the exponent becomes $k$ .
4
Combine coefficients and match the target.
$\begin{aligned} \sum_{r = 1}^{k + 1} r 2^{r - 1} & = 2 k 2^{k} + 1 \\ = k 2^{k + 1} + 1 \\ = ((k + 1) - 1) 2^{k + 1} + 1. \end{aligned}$
The coefficient sum is $(k - 1) + (k + 1) = 2 k$ .
5
Finish the proof.
$\sum_{r = 1}^{n} r 2^{r - 1} = (n - 1) 2^{n} + 1 (n \in ℤ, n \geq 1) .$
The last expression is the next case, so induction establishes all positive integers.

Answer

$\sum_{r = 1}^{n} r 2^{r - 1} = (n - 1) 2^{n} + 1$ is true throughout the stated domain.

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$3^{k}$ .

The first $k$ terms have exponents $0, 1, \dots, k - 1$ . The next exponent is $k$ , so the next term is $3^{k}$ .

Keep practising

6 worksheet questions · 36 marks

Open worksheet 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 $1 + 3 + 3^{2} + \dots + 3^{n - 1} = \frac{3^{n} - 1}{2}$ 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 $\prod_{r = 1}^{n} \frac{r + 1}{r} = n + 1$ 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 $\prod_{r = 1}^{n} \frac{2 r - 1}{2 r + 1} = \frac{1}{2 n + 1}$ 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 $\sum_{r = 1}^{n} {(2 r - 1)}^{2} = \frac{n (4 n^{2} - 1)}{3}$ 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 $1 + 3 + 3^{2} + \dots + 3^{n - 1} = \frac{3^{n} - 1}{2}$ for all integers $n \geq 1$ .
6 marks
(a) $n = 1$ gives $\frac{3 - 1}{2} = 1$ .
(a) Assume the sum to $k$ is $\frac{3^{k} - 1}{2}$ .
(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 $\frac{3 - 1}{2} = 1$ .
  Assume the sum to $k$ is $\frac{3^{k} - 1}{2}$ .
  $n = 1$ gives $\frac{3 - 1}{2} = 1$ . Assume the sum to $k$ is $\frac{3^{k} - 1}{2}$ .
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 $\frac{3 - 1}{2} = 1$ . Assume the sum to $k$ is $\frac{3^{k} - 1}{2}$ . (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 $\prod_{r = 1}^{n} \frac{r + 1}{r} = n + 1$ 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 $\frac{k + 2}{k + 1}$ .
(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 $\frac{k + 2}{k + 1}$ .
  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 $\prod_{r = 1}^{n} \frac{2 r - 1}{2 r + 1} = \frac{1}{2 n + 1}$ for all integers $n \geq 1$ .
6 marks
(a) At $n = 1$ , the product is $\frac{1}{3}$ .
(a) Assume the product to $k$ is $\frac{1}{2 k + 1}$ .
(b) Multiply by $\frac{2 k + 1}{2 k + 3}$ .
(b) The result is $\frac{1}{2 k + 3} = \frac{1}{2 (k + 1) + 1}$ .
(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 $\frac{1}{3}$ .
  Assume the product to $k$ is $\frac{1}{2 k + 1}$ .
  At $n = 1$ , the product is $\frac{1}{3}$ . Assume the product to $k$ is $\frac{1}{2 k + 1}$ .
2. (b)
  Prove the next case and conclude.
  4 marks
  Multiply by $\frac{2 k + 1}{2 k + 3}$ .
  The result is $\frac{1}{2 k + 3} = \frac{1}{2 (k + 1) + 1}$ .
  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 $\frac{1}{3}$ . Assume the product to $k$ is $\frac{1}{2 k + 1}$ . (b) The statement holds for all positive integers.
Prove $\sum_{r = 1}^{n} {(2 r - 1)}^{2} = \frac{n (4 n^{2} - 1)}{3}$ 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.

Official curriculum source ↗

In this lesson

Understand the idea.

Track the exponent

Match the operation

See the method in action.

A geometric sum

Verify the one-term sum.

Assume the first k terms.

Append the next power.

Combine the common power.

Conclude the identity.

A weighted power identity

Check the first weighted term.

Assume the identity at k.

Add the new weighted term.

Combine coefficients and match the target.

Finish the proof.

Try it yourself.

Check your understanding

Keep practising

Assume the first $k$ terms.

Assume the identity at $k$ .