MATHEMATICAL INDUCTION AND THE BINOMIAL THEOREM: FORMULAS, PROOFS, EXPANSIONS AND SOLVED EXAMPLES
The Target-First Domino Method and the Pick-the-Brackets Shortcut
Mathematical induction and the Binomial Theorem solve two very different mathematical problems.
Mathematical induction proves that a statement remains true for every integer in an infinite sequence. The Binomial Theorem expands expressions such as (x + 2)⁸ without multiplying eight brackets individually.
Yet these two topics have a surprisingly useful connection. Pascal's triangle, combinations and mathematical induction work together to explain why binomial coefficients follow their familiar patterns.
This guide explains mathematical induction proofs, algebraic sum formulas, divisibility arguments, inequality proofs, Pascal's triangle, binomial expansion, general terms, specific coefficients and constant-term questions.
It is useful for students studying Algebra 2, Precalculus, advanced high-school mathematics, discrete mathematics, AP-related mathematics, UK A-Level Mathematics and Further Mathematics, and international mathematics programmes where these topics are included.
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PART 1: MATHEMATICAL INDUCTION
1. Why Checking Examples Is Not Mathematical Proof
Consider the expression:
n² + n + 41
Substitute different non-negative integers for n.
n = 0: 0² + 0 + 41 = 41
n = 1: 1² + 1 + 41 = 43
n = 2: 2² + 2 + 41 = 47
n = 3: 3² + 3 + 41 = 53
n = 4: 4² + 4 + 41 = 61
The expression continues producing prime numbers for every integer from 0 through 39.
It looks as though we have discovered a remarkable mathematical rule.
Now substitute n = 40.
40² + 40 + 41
= 1600 + 40 + 41
= 1681
= 41 × 41
The result is not prime.
Forty successful examples did not establish a universal mathematical statement.
This illustrates an important distinction:
• Observation: A pattern appears to work in the examples examined.
• Conjecture: We suspect that the pattern always works.
• Proof: We establish logically that the statement must be true for every value in its specified domain.
Mathematical induction is one of the most important methods for proving statements involving positive integers or non-negative integers.
It is particularly useful for:
• Proving algebraic sum formulas.
• Proving divisibility rules.
• Establishing inequalities involving integers.
• Proving identities involving powers.
• Establishing properties of sequences.
• Proving formulas involving factorials and binomial coefficients.
2. The Domino Principle: Understanding Mathematical Induction
Imagine a row of dominoes numbered 1, 2, 3, 4 and so on.
To establish that every domino falls, two things must happen.
Condition 1: The first domino falls.
This is the base case.
Condition 2: Every falling domino knocks over the next domino.
This is the inductive step.
1 → 2 → 3 → 4 → 5 → …
If the first domino falls and each domino causes the next one to fall, the entire sequence follows.
Mathematical induction uses exactly this reasoning.
Suppose P(n) represents a mathematical statement involving the integer n.
To prove the statement for every integer n ≥ 1, establish:
Step 1: Base case
Show that P(1) is true.
Step 2: Inductive hypothesis
Assume that P(k) is true for an arbitrary integer k ≥ 1.
This assumption is temporary. It is made to establish the next case.
Step 3: Inductive step
Using the assumption P(k), prove that P(k + 1) is true.
Step 4: Conclusion
Since the starting case is true and each case implies the next one, the statement holds for every integer n ≥ 1.
The starting value is not always 1. If a question asks you to prove a statement for n ≥ 5, the base case must begin at 5.
3. The Target-First Domino Method
One of the biggest difficulties in mathematical induction is knowing where to begin the algebra.
Students often write several lines of manipulation without knowing what expression they are trying to obtain.
A more reliable approach is to write the destination first.
Use this five-stage method.
Stage 1: State the proposition
Write the mathematical statement as P(n).
Stage 2: Verify the base case
Substitute the starting value and check that both sides agree, or that the required inequality holds.
Stage 3: Assume the kth case
Write P(k) explicitly.
Do not assume the result for k + 1. That is the very statement you are supposed to prove.
Stage 4: Write the target
Replace every occurrence of n with k + 1.
This gives the exact expression you need to establish.
Stage 5: Build the bridge
Start with the expression in the target, identify the part corresponding to P(k), substitute the inductive hypothesis and simplify.
Finish with a conclusion.
This method works especially well for sums, divisibility and many algebraic identities.
4. Mathematical Induction Example 1: Proving the Sum of the First n Integers
Prove that:
1 + 2 + 3 + … + n = n(n + 1)/2
for every integer n ≥ 1.
Step 1: State the proposition
P(n): 1 + 2 + … + n = n(n + 1)/2
Step 2: Base case
Let n = 1.
Left-hand side = 1
Right-hand side = 1(1 + 1)/2 = 1
Both sides are equal. The base case is true.
Step 3: Inductive hypothesis
Assume the formula is true for n = k:
1 + 2 + … + k = k(k + 1)/2
Step 4: Target
We must prove:
1 + 2 + … + k + (k + 1) = (k + 1)(k + 2)/2
Step 5: Build the bridge
Begin with the left-hand side:
1 + 2 + … + k + (k + 1)
Use the inductive hypothesis:
= k(k + 1)/2 + (k + 1)
Factor out the common term:
= (k + 1)(k/2 + 1)
Simplify:
= (k + 1)(k + 2)/2
This is exactly the required result.
Conclusion: By mathematical induction,
1 + 2 + 3 + … + n = n(n + 1)/2
for every integer n ≥ 1.
The important idea is to separate the newest term, k + 1, from the existing sum. Everything before it becomes the expression covered by the inductive hypothesis.
5. Mathematical Induction Example 2: Proving the Sum of Consecutive Odd Numbers
Prove that:
1 + 3 + 5 + … + (2n − 1) = n²
for every integer n ≥ 1.
Base case:
For n = 1:
1 = 1²
True.
Inductive hypothesis:
Assume:
1 + 3 + … + (2k − 1) = k²
Target:
For k + 1, the final odd number becomes:
2(k + 1) − 1 = 2k + 1
We must prove:
1 + 3 + … + (2k − 1) + (2k + 1) = (k + 1)²
Inductive step:
Using the assumption:
= k² + 2k + 1
Factor:
= (k + 1)²
The target is established.
Conclusion: The sum of the first n positive odd integers is n².
This example shows that induction can prove formulas for sequences whose terms are not consecutive integers.
6. Mathematical Induction Example 3: Divisibility Proofs
Prove that:
7ⁿ − 1
is divisible by 6 for every integer n ≥ 1.
A divisibility statement means that the expression can be written as an integer multiple of the required divisor.
Base case:
For n = 1:
7¹ − 1 = 6
Divisible by 6.
Inductive hypothesis:
Assume that:
7ᵏ − 1 = 6m
where m is an integer.
Target:
Show that:
7ᵏ⁺¹ − 1
is divisible by 6.
Inductive step:
Rewrite the expression:
7ᵏ⁺¹ − 1 = 7 × 7ᵏ − 1
Add and subtract 7:
= 7(7ᵏ − 1) + 6
Use the inductive hypothesis:
= 7(6m) + 6
Factor out 6:
= 6(7m + 1)
Since 7m + 1 is an integer, the expression is divisible by 6.
Conclusion: By mathematical induction, 7ⁿ − 1 is divisible by 6 for every integer n ≥ 1.
Divisibility shortcut: Try to rewrite the new expression as a multiple of the assumed expression plus another term that is divisible by the required number.
7. Mathematical Induction Example 4: Proving an Inequality
Prove that:
2ⁿ > n²
for every integer n ≥ 5.
Notice the starting value.
At n = 3:
2³ = 8 < 9 = 3²
At n = 4:
2⁴ = 16 = 4²
Therefore, the statement is not true for all positive integers. It only becomes true from n = 5 onwards.
Base case:
For n = 5:
2⁵ = 32 > 25 = 5²
True.
Inductive hypothesis:
Assume that for an integer k ≥ 5:
2ᵏ > k²
Target:
Prove:
2ᵏ⁺¹ > (k + 1)²
Inductive step:
Start with the left-hand side:
2ᵏ⁺¹ = 2 × 2ᵏ
Using the assumption:
2ᵏ⁺¹ > 2k²
It remains to establish:
2k² ≥ (k + 1)²
Expand the right-hand side:
2k² ≥ k² + 2k + 1
Rearrange:
k² − 2k − 1 ≥ 0
Since k ≥ 5:
k² ≥ 5k
Therefore:
k² − 2k − 1 ≥ 3k − 1 > 0
Hence:
2ᵏ⁺¹ > 2k² ≥ (k + 1)²
The required inequality follows.
Conclusion: By mathematical induction,
2ⁿ > n² for every integer n ≥ 5.
The lesson is important: an inequality proof may require an additional comparison after applying the inductive hypothesis.
8. Why the Base Case Cannot Be Skipped
Consider the false statement:
n² + n is odd for every positive integer n.
Suppose we assume that:
k² + k
is odd.
For the next integer:
(k + 1)² + (k + 1)
Expand:
= k² + 2k + 1 + k + 1
Rearrange:
= (k² + k) + (2k + 2)
The first part is assumed to be odd. The second part is even.
Odd plus even is odd.
The inductive step works.
But test the starting value:
1² + 1 = 2
which is even.
The original statement is false.
The inductive step only tells us that truth passes from one case to the next. It does not tell us that the chain ever begins.
This is why a mathematical induction proof is incomplete without its base case.
9. Strong Mathematical Induction: When One Previous Case Is Not Enough
Ordinary mathematical induction assumes that P(k) is true and proves P(k + 1).
Strong mathematical induction uses a broader assumption.
Instead of assuming only P(k), we assume that all the earlier statements are true:
P(1), P(2), P(3), …, P(k)
We then prove P(k + 1).
This is useful when the next case depends on more than one earlier case.
For example, consider the Fibonacci sequence:
F₁ = 1, F₂ = 1
Fₙ₊₂ = Fₙ₊₁ + Fₙ
To establish a property of this sequence, we often need two starting cases and information about two preceding terms.
Strong induction is also useful in divisibility, integer factorisation and number theory.
Both ordinary and strong induction are valid proof techniques. The appropriate method depends on how the statement is constructed.
10. Five Common Mathematical Induction Mistakes
Skipping the base case: The logical chain has no starting point.
Assuming the next case: Assuming P(k + 1) instead of P(k) is circular reasoning.
Failing to use the assumption: The inductive hypothesis must actually contribute to the proof.
Using an incorrect starting value: Inequalities may only become true after a particular integer.
Omitting the conclusion: Explicitly state that the statement is true for all integers in the specified range.
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PART 2: THE BINOMIAL THEOREM
11. What Is a Binomial Expression?
A binomial is an algebraic expression containing two terms.
Examples include:
(x + 2), (a + b), (3x − 5)
When a binomial is raised to a power, expanding it by repeated multiplication becomes increasingly inconvenient.
For example:
(x + 2)² = (x + 2)(x + 2)
= x² + 4x + 4
Similarly:
(x + 2)³ = x³ + 6x² + 12x + 8
The Binomial Theorem provides a general formula that allows us to expand any non-negative integer power without multiplying every bracket individually.
12. Discovering the Pattern Through Small Expansions
Start with the simplest examples.
(a + b)⁰ = 1
(a + b)¹ = a + b
(a + b)² = a² + 2ab + b²
(a + b)³ = a³ + 3a²b + 3ab² + b³
(a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
Three patterns appear.
Pattern 1: Powers of the first term decrease.
a⁴, a³, a², a, 1
Pattern 2: Powers of the second term increase.
1, b, b², b³, b⁴
Pattern 3: The total power in each term remains constant.
For example:
a²b²
has a total power of 4.
The coefficients follow the pattern:
1, 4, 6, 4, 1
These numbers come from Pascal's triangle.
13. Pascal's Triangle and Binomial Coefficients
Pascal's triangle is a triangular arrangement of numbers in which each interior number is the sum of the two numbers immediately above it.
Row 0: 1
Row 1: 1 1
Row 2: 1 2 1
Row 3: 1 3 3 1
Row 4: 1 4 6 4 1
Row 5: 1 5 10 10 5 1
Row 6: 1 6 15 20 15 6 1
The first row is numbered 0.
Therefore:
Power 0: 1
Power 1: 1, 1
Power 2: 1, 2, 1
Power 3: 1, 3, 3, 1
Power 4: 1, 4, 6, 4, 1
Power 5: 1, 5, 10, 10, 5, 1
Power 6: 1, 6, 15, 20, 15, 6, 1
The row numbered n provides the coefficients for expanding (a + b)ⁿ.
This is convenient for smaller powers. For larger powers, we need the general formula.
14. The Pick-the-Brackets Shortcut: Why the Binomial Theorem Works
Imagine expanding:
(a + b)⁴
Write it as four identical brackets:
(a + b)(a + b)(a + b)(a + b)
To construct a term, choose either a or b from every bracket.
For example, choosing a three times and b once gives:
a³b
But how many different ways can we make this selection?
There are four possible brackets from which to select b.
Therefore, the coefficient of a³b is 4.
If we choose b twice, we get:
a²b²
There are six ways to select two of the four brackets.
Therefore, the coefficient is 6.
This is the fundamental reason binomial coefficients appear in the expansion.
The number of ways of choosing r objects from n objects is called a combination.
It is written as:
ⁿCᵣ or C(n, r)
It is also written as nCr on many calculators.
The formula is:
ⁿCᵣ = n! / [r!(n − r)!]
Here, the factorial symbol means:
n! = n(n − 1)(n − 2) … 2 × 1
For example:
⁵C₂ = 5! / (2! × 3!)
= 120 / (2 × 6)
= 10
There are 10 ways to select two objects from five.
15. The Binomial Theorem: The General Formula
For a non-negative integer n:
(a + b)ⁿ = Σ ⁿCᵣ aⁿ⁻ʳbʳ
where r runs from 0 to n.
Written without summation notation:
(a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + … + ⁿCₙbⁿ
Every term follows the same structure:
• The coefficient comes from a combination.
• The power of a decreases.
• The power of b increases.
• The sum of the two powers is always n.
The general term is:
Tᵣ₊₁ = ⁿCᵣ × aⁿ⁻ʳ × bʳ
The term number is r + 1, not r, because the first term corresponds to r = 0.
This small distinction is responsible for many examination mistakes.
16. The Three-Slot Check
Before accepting any binomial expansion, check three things.
Slot 1: Coefficient
Is the coefficient the correct combination?
ⁿCᵣ
Slot 2: First power
Does the power of the first term decrease correctly?
n, n − 1, n − 2, …, 0
Slot 3: Second power
Does the power of the second term increase correctly?
0, 1, 2, …, n
Finally, check that the two powers in every term add up to n.
The coefficients should also be symmetrical.
For example:
1, 5, 10, 10, 5, 1
This provides a quick way to identify mistakes.
17. Worked Example: Expanding (x + 2)⁴
Expand:
(x + 2)⁴
The fourth row of Pascal's triangle is:
1, 4, 6, 4, 1
Substitute a = x and b = 2:
(x + 2)⁴
= x⁴ + 4x³(2) + 6x²(2²) + 4x(2³) + 2⁴
Simplify:
= x⁴ + 8x³ + 24x² + 32x + 16
Therefore:
x⁴ + 8x³ + 24x² + 32x + 16
Check the result.
Set x = 1.
Original expression:
(1 + 2)⁴ = 81
Expanded expression:
1 + 8 + 24 + 32 + 16 = 81
Both agree.
This substitution is a useful numerical check, although it is not a substitute for an algebraic proof.
18. Worked Example: Expanding a Binomial with a Negative Term
Expand:
(2x − 3)³
Treat the expression as:
(2x + (−3))³
The third row of Pascal's triangle is:
1, 3, 3, 1
Apply the theorem:
(2x)³ + 3(2x)²(−3) + 3(2x)(−3)² + (−3)³
Simplify:
= 8x³ − 36x² + 54x − 27
Therefore:
8x³ − 36x² + 54x − 27
Important sign rule: Treat the negative term as a complete quantity inside brackets. Its sign must be raised to the relevant power.
For a binomial of the form (a − b)ⁿ, the signs alternate, beginning with a positive first term.
19. How to Find a Particular Term in a Binomial Expansion
Sometimes an examination question does not ask you to expand the entire expression.
Instead, it asks for:
• The coefficient of a particular power of x.
• The fourth or fifth term.
• A specified term in an expansion.
• The coefficient of a particular algebraic expression.
The general-term formula makes these questions much easier.
Consider:
Find the coefficient of x³ in:
(2x + 1)⁷
Start with:
Tᵣ₊₁ = ⁷Cᵣ(2x)⁷⁻ʳ(1)ʳ
The power of x is:
7 − r
We need the power to equal 3.
Therefore:
7 − r = 3
r = 4
Substitute:
T₅ = ⁷C₄(2x)³(1)⁴
= 35 × 8x³
= 280x³
Therefore, the coefficient is:
280
The method to remember:
Write the general term.
Identify the power of x.
Equate that power to the one required.
Solve for r.
Substitute the value into the general term.
This method avoids expanding terms that the question never asks for.
20. How to Find the Constant Term in a Binomial Expansion
The constant term is the term that does not contain the variable.
Its power of x is zero.
Consider:
Find the constant term in:
(x² + 1/x)⁹
First rewrite the fraction:
1/x = x⁻¹
The general term is:
Tᵣ₊₁ = ⁹Cᵣ(x²)⁹⁻ʳ(x⁻¹)ʳ
Apply the laws of indices:
Tᵣ₊₁ = ⁹Cᵣ x¹⁸⁻²ʳ⁻ʳ
Therefore:
Tᵣ₊₁ = ⁹Cᵣ x¹⁸⁻³ʳ
For the constant term:
18 − 3r = 0
3r = 18
r = 6
The required coefficient is:
⁹C₆ = 9! / (6! × 3!) = 84
Therefore:
Constant term = 84
A useful shortcut: In constant-term questions, convert fractions into negative powers first. Then find the value of r that makes the exponent zero.
If the resulting value of r is not an integer between 0 and n, there is no constant term in the expansion.
21. Binomial Expansions Involving Negative Powers of x
Consider:
(x + 2/x)⁶
The general term is:
Tᵣ₊₁ = ⁶Cᵣ x⁶⁻ʳ(2/x)ʳ
Simplify:
Tᵣ₊₁ = ⁶Cᵣ 2ʳx⁶⁻²ʳ
For the constant term:
6 − 2r = 0
2r = 6
r = 3
Therefore:
T₄ = ⁶C₃ × 2³
= 20 × 8
= 160
Notice that the constant term is the fourth term, not the third term.
The position of a term depends on the value of r + 1.
22. How to Find the Sum of Coefficients in a Binomial Expansion
There is a particularly useful shortcut for finding the sum of all coefficients.
Suppose an expression is:
P(x) = a₀ + a₁x + a₂x² + … + aₙxⁿ
Substitute:
x = 1
Then:
P(1) = a₀ + a₁ + a₂ + … + aₙ
This is exactly the sum of the coefficients.
Worked example
Find the sum of the coefficients in:
(3x − 1)⁶
Set x = 1:
(3(1) − 1)⁶
= 2⁶
= 64
Similarly, the sum of the coefficients in:
(1 + x)ⁿ
is:
2ⁿ
This also explains why the row sums of Pascal's triangle are powers of 2.
23. The Alternating Sum of Binomial Coefficients
Another useful identity comes from substituting x = −1.
For the expansion:
(1 + x)ⁿ
put x = −1:
(1 − 1)ⁿ = 0
For n ≥ 1:
ⁿC₀ − ⁿC₁ + ⁿC₂ − … + (−1)ⁿⁿCₙ = 0
This means that the sum of the even-indexed binomial coefficients equals the sum of the odd-indexed binomial coefficients.
For example:
1 − 5 + 10 − 10 + 5 − 1 = 0
This identity is useful in questions involving alternating binomial coefficients.
24. Binomial Coefficients and Combinations
The Binomial Theorem is closely related to combinations in probability.
Suppose a coin is tossed five times.
The number of possible ways to obtain exactly two heads is:
⁵C₂ = 10
The same combination appears as the coefficient of the corresponding term in:
(a + b)⁵
Why?
Each of the five tosses has two possible outcomes. Choosing which two positions contain heads is exactly the same selection process as choosing which two brackets contribute the second term.
This connection explains why binomial coefficients appear in:
• Probability distributions.
• Counting arrangements.
• Combinatorics.
• Algebraic expansions.
• Pascal's triangle.
25. A Practical Application: Compound Interest and Binomial Expansion
The Binomial Theorem can also help estimate compound growth.
Suppose $1 or £1 is invested at an annual interest rate of 2%, compounded annually for five years.
The future value of one unit of currency is:
(1 + 0.02)⁵
Using the Binomial Theorem:
= 1 + 5(0.02) + 10(0.02)² + 10(0.02)³ + 5(0.02)⁴ + (0.02)⁵
Calculate:
= 1 + 0.1 + 0.004 + 0.00008 + 0.0000008 + 0.0000000032
Therefore:
= 1.1040808032
The investment grows to approximately:
$1.10408 or £1.10408
for every original dollar or pound invested.
This is not a replacement for a financial calculator. It demonstrates how algebraic expansion can explain the structure of compound growth.
26. The Connection Between Mathematical Induction and the Binomial Theorem
The Binomial Theorem can itself be proved using mathematical induction.
Recall the formula:
(a + b)ⁿ = Σ ⁿCᵣaⁿ⁻ʳbʳ
We can establish it by proving that it works for the first case and then showing that if it works for n = k, it must also work for n = k + 1.
Base case:
For n = 0:
(a + b)⁰ = 1
The right-hand side contains only:
⁰C₀a⁰b⁰ = 1
Therefore, the base case is true.
Inductive hypothesis:
Assume:
(a + b)ᵏ = Σ ⁿCᵣaᵏ⁻ʳbʳ
where the intended upper index is k and the summation runs from r = 0 to k.
Inductive step:
Multiply both sides by (a + b):
(a + b)ᵏ⁺¹ = (a + b) × Σ ᵏCᵣaᵏ⁻ʳbʳ
Distribute:
= Σ ᵏCᵣaᵏ⁺¹⁻ʳbʳ + Σ ᵏCᵣaᵏ⁻ʳbʳ⁺¹
Now collect terms having the same powers of a and b.
The coefficient of the general interior term becomes:
ᵏCᵣ + ᵏCᵣ₋₁
Pascal's identity tells us:
ᵏCᵣ + ᵏCᵣ₋₁ = ᵏ⁺¹Cᵣ
Consequently:
(a + b)ᵏ⁺¹ = Σ ᵏ⁺¹Cᵣaᵏ⁺¹⁻ʳbʳ
This is exactly the Binomial Theorem for k + 1.
The base case and inductive step are established, so the theorem holds for every non-negative integer n.
The connection is worth remembering:
Mathematical induction proves that the Binomial Theorem works, while Pascal's identity explains how the coefficients develop from one power to the next.
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PART 3: EXTENDING THE BINOMIAL THEOREM
27. What Happens with Negative and Fractional Powers?
The ordinary Binomial Theorem applies to non-negative integer powers.
However, there is an extended version that allows expressions such as:
(1 + x)⁻¹
and:
(1 + x)¹ᐟ²
These expansions generally produce infinite series rather than finite polynomials.
The generalized binomial expansion is:
(1 + x)ʳ = 1 + rx + [r(r − 1)/2!]x² + [r(r − 1)(r − 2)/3!]x³ + …
For a general negative or non-integer real exponent r, this standard power series converges when:
|x| < 1
Unlike an ordinary finite expansion, the series does not usually terminate.
Example 1: Negative power
For:
(1 + x)⁻¹
the expansion is:
1 − x + x² − x³ + x⁴ − …
This is the familiar geometric series, valid when |x| < 1.
Example 2: Fractional power
For:
(1 + x)¹ᐟ²
the expansion begins:
1 + x/2 − x²/8 + x³/16 − 5x⁴/128 + …
These extended expansions are important in advanced algebra, A-Level Mathematics and calculus.
They also provide a bridge between finite algebraic expansions and infinite power series.
28. When Should You Use Pascal's Triangle and When Should You Use the Formula?
Question type | Best method
Expand (a + b)³ | Pascal's triangle
Expand (x + 2)⁵ | Pascal's triangle or formula
Expand (2x − 3)⁸ | General formula
Find the sixth term | General-term formula
Find the coefficient of x⁴ | General-term formula
Find the constant term | General-term formula
Find the sum of coefficients | Substitute x = 1
Find alternating coefficient sum | Substitute x = −1
Expand a negative integer power | Generalized binomial expansion
Expand a fractional power | Generalized binomial expansion
The most efficient method depends on what the question actually asks.
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PART 4: EXAMINATION STRATEGY AND PRACTICE
29. Common Binomial Theorem Examination Mistakes
Incorrect term numbering: The general term is Tᵣ₊₁, not Tᵣ.
Forgetting to raise the coefficient: (2x)³ = 8x³, not 2x³.
Losing negative signs: Keep the negative term inside brackets throughout the calculation.
Incorrect powers: The powers of the two terms must add to n.
Confusing a term with its coefficient: If the term is 280x³, the coefficient is 280.
Ignoring negative indices: Rewrite 1/x as x⁻¹ before simplifying.
Expanding unnecessarily: If the question asks for one term, use the general-term formula.
Using the finite theorem incorrectly: Negative and fractional powers require the generalized expansion.
30. Practice Questions: Mathematical Induction
Try the following questions without looking at the answers.
Question 1
Prove by mathematical induction that:
1 + 2 + 3 + … + n = n(n + 1)/2
for all integers n ≥ 1.
Question 2
Prove that:
1 + 3 + 5 + … + (2n − 1) = n²
for all integers n ≥ 1.
Question 3
Prove by induction that:
5ⁿ − 1
is divisible by 4 for all integers n ≥ 1.
Question 4
Examine the statement:
3ⁿ > 2n + 1
for all integers n ≥ 1.
Determine whether it is true. If not, identify the appropriate starting value and prove the corrected statement.
Question 5
Prove that:
2ⁿ ≥ n + 1
for all non-negative integers n.
31. Practice Questions: The Binomial Theorem
Question 6
Expand:
(x − 3)⁴
Question 7
Expand:
(2x + 1)⁵
Question 8
Find the coefficient of x² in:
(1 + 2x)⁶
Question 9
Find the coefficient of x³ in:
(2x + 1)⁷
Question 10
Find the constant term in:
(x + 2/x)⁶
Question 11
Find the constant term in:
(x² + 1/x)⁹
Question 12
Find the sum of the coefficients in:
(2x + 1)⁵
Question 13
Find the alternating sum of the coefficients in:
(1 + x)⁸
Question 14
Find the fifth term in:
(3x − 2)⁷
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PART 5: ANSWERS AND EXPLANATIONS
Mathematical Induction Answers
Answer 1
Base case:
n = 1:
1 = 1(2)/2
Assume:
1 + 2 + … + k = k(k + 1)/2
Then:
1 + 2 + … + k + (k + 1)
= k(k + 1)/2 + (k + 1)
= (k + 1)(k + 2)/2
Therefore, the formula holds for every integer n ≥ 1.
Answer 2
Base case:
1 = 1²
Assume:
1 + 3 + … + (2k − 1) = k²
Add the next odd number:
k² + (2k + 1)
= k² + 2k + 1
= (k + 1)²
The statement follows by induction.
Answer 3
Base case:
5¹ − 1 = 4
Assume:
5ᵏ − 1 = 4m
Then:
5ᵏ⁺¹ − 1 = 5(5ᵏ − 1) + 4
= 5(4m) + 4
= 4(5m + 1)
Therefore, the expression is divisible by 4.
Answer 4
The statement is false at n = 1:
3¹ = 3
2(1) + 1 = 3
The two sides are equal, not strictly greater.
The corrected statement is:
3ⁿ > 2n + 1 for all integers n ≥ 2.
Base case:
3² = 9 > 5
Assume:
3ᵏ > 2k + 1
Then:
3ᵏ⁺¹ > 3(2k + 1)
= 6k + 3
For k ≥ 2:
6k + 3 > 2k + 3
Therefore:
3ᵏ⁺¹ > 2(k + 1) + 1
Hence the statement holds for every integer n ≥ 2.
Answer 5
Base case:
2⁰ = 1 = 0 + 1
Assume:
2ᵏ ≥ k + 1
Then:
2ᵏ⁺¹ = 2 × 2ᵏ
≥ 2(k + 1)
Since k ≥ 0:
2k + 2 ≥ k + 2
Therefore:
2ᵏ⁺¹ ≥ (k + 1) + 1
The statement holds for every non-negative integer n.
Binomial Theorem Answers
Answer 6
(x − 3)⁴
= x⁴ − 12x³ + 54x² − 108x + 81
Answer 7
(2x + 1)⁵
= 32x⁵ + 80x⁴ + 80x³ + 40x² + 10x + 1
Answer 8
General term:
Tᵣ₊₁ = ⁶Cᵣ(1)⁶⁻ʳ(2x)ʳ
For x², r = 2.
⁶C₂ × 2²
= 15 × 4
= 60
Answer 9
Tᵣ₊₁ = ⁷Cᵣ(2x)⁷⁻ʳ
For x³:
7 − r = 3
r = 4
⁷C₄ × 2³
= 35 × 8
= 280
Answer 10
Tᵣ₊₁ = ⁶Cᵣ × 2ʳ × x⁶⁻²ʳ
For a constant:
6 − 2r = 0
r = 3
⁶C₃ × 2³
= 20 × 8
= 160
Answer 11
Tᵣ₊₁ = ⁹Cᵣ × x¹⁸⁻³ʳ
For a constant:
18 − 3r = 0
r = 6
⁹C₆ = 84
Answer 12
Set x = 1:
(2 + 1)⁵ = 3⁵
= 243
Answer 13
Set x = −1:
(1 − 1)⁸ = 0
Answer 14
The fifth term corresponds to r = 4.
T₅ = ⁷C₄(3x)³(−2)⁴
= 35 × 27x³ × 16
= 15,120x³
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THE FINAL TWO IDEAS TO REMEMBER
Mathematical induction: Establish the starting case, assume the kth case, use that assumption to reach the next case, and complete the proof with a clear conclusion.
The Binomial Theorem: Every term is formed by choosing the second term from a certain number of brackets. The number of ways to make that selection determines the coefficient.
These are not merely two collections of examination formulas. They are examples of how mathematics uses logical structure to solve problems that would otherwise require endless checking or repetitive multiplication.
Understand the reasoning behind the methods, and the formulas become much easier to remember and apply.
