Mathematical induction and binomial theorem: Class 11 Mathematics Notes, Short Questions and Numericals

The principle of mathematical induction with proofs of sums and divisibility, the binomial theorem for a positive integer index, the general term, middle terms, Pascal's triangle and the binomial series for approximations.

Key points

Mathematical induction is a method of proving that a statement P(n) is true for every natural number n. A pattern seen in a few cases is not a proof, so induction gives a reliable way to prove it for all n. It works like a row of dominoes: if the first falls, and each one knocks down the next, then all fall.

Steps of a proof by induction:

Basis step: show that P(1) is true (or the first value of n given).
Induction hypothesis: assume P(k) is true for some natural number k.
Induction step: using this assumption, prove that P(k + 1) is true.
Conclusion: by the principle of mathematical induction, P(n) is true for all natural numbers n.

Induction is often used for sums. To prove 1 + 2 + … + n = n(n + 1)/2, check n = 1, assume the result for k, and add the next term k + 1 to both sides. The right side becomes k(k + 1)/2 + (k + 1) = (k + 1)(k + 2)/2, which is the formula with n = k + 1. It is also used for divisibility, for example that 7ⁿ − 1 is divisible by 6, and for inequalities such as 2ⁿ > n.

For divisibility, write the expression for k + 1 so that it contains the expression for k. For 7ⁿ − 1: 7ᵏ⁺¹ − 1 = 7(7ᵏ − 1) + 6. The first part is divisible by 6 by the assumption, and the second part is 6, so the sum is divisible by 6.

A binomial is an expression with two terms, such as (a + b). The binomial theorem gives the expansion of (a + b)ⁿ for a positive integer n: (a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + … + ⁿCₙbⁿ. The numbers ⁿCᵣ are the binomial coefficients. The expansion has n + 1 terms. In each term the powers of a and b add up to n, and the powers of a go down while the powers of b go up.

The coefficients can be read from Pascal's triangle, in which each number is the sum of the two above it. The rows are 1; 1 1; 1 2 1; 1 3 3 1; 1 4 6 4 1; and so on. The row for n = 4 gives the coefficients of (a + b)⁴. The coefficients are symmetric, since ⁿCᵣ = ⁿCₙ₋ᵣ. For (a − b)ⁿ, the signs alternate: +, −, +, −, and so on.

The general term, or (r + 1)th term, is Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ. It is used to find a particular term without expanding the whole expression. For example the 6th term uses r = 5. To find the term containing a given power of x, write the power of x in the general term, equate it to the required power, and solve for r. The term independent of x is the one in which the power of x is 0.

The middle term depends on n. If n is even, there is one middle term, the (n/2 + 1)th term. If n is odd, there are two middle terms, the ((n + 1)/2)th and ((n + 3)/2)th terms. The sum of all binomial coefficients is 2ⁿ, obtained by putting a = b = 1. Putting a = 1 and b = −1 shows that the sum of the even placed coefficients equals the sum of the odd placed ones.

When x is small, the binomial theorem gives quick approximations: (1 + x)ⁿ ≈ 1 + nx + n(n − 1)x²/2 + …, and the terms become smaller as x is small. This is how (1.02)⁴ is estimated by taking x = 0.02. The same idea extends to negative and fractional n in an infinite series that holds when |x| < 1.

Study tip: In an induction proof, write the statement P(k + 1) clearly before you begin and work towards it. For binomial questions, always begin with the general term, simplify the power of x, and then solve for r. Check that r is a whole number from 0 to n.

Definitions and flashcards

Say the meaning first, then open the card to check.

Mathematical induction

A proof method for statements about all natural numbers.

Induction hypothesis

The assumption that the statement is true for n = k.

Binomial

An algebraic expression with two terms.

Binomial coefficient

The number ⁿCᵣ that multiplies aⁿ⁻ʳbʳ in the expansion.

Pascal's triangle

A triangle of numbers in which each entry is the sum of the two above it.

General term

The formula Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ for the (r + 1)th term.

Middle term

The term in the centre of the expansion, one term if n is even and two if n is odd.

Formula sheet

Name	Formula	What it means
Binomial theorem	(a + b)ⁿ = Σ ⁿCᵣ aⁿ⁻ʳ bʳ, r = 0 to n	n is a positive integer.
General term	Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ	The (r + 1)th term.
Number of terms	n + 1	In the expansion of (a + b)ⁿ.
Sum of coefficients	ⁿC₀ + ⁿC₁ + … + ⁿCₙ = 2ⁿ	Put a = b = 1.
Small x approximation	(1 + x)ⁿ ≈ 1 + nx + n(n − 1)x²/2	Use when x is small.

Short questions with model answers

These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.

List the steps of the principle of mathematical induction. (2 marks)

First show that the statement is true for n = 1. Then assume that it is true for n = k. Then prove that it is true for n = k + 1 using this assumption. Finally conclude that the statement is true for every natural number n.

How many terms are there in the expansion of (a + b)ⁿ? Write the general term. (2 marks)

There are n + 1 terms. The general term, which is the (r + 1)th term, is Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ.

Find the middle term in the expansion of (x + 1)⁶. (3 marks)

n = 6 is even, so there is one middle term, the (6/2 + 1)th = 4th term. T₄ = ⁶C₃ x³(1)³ = 20x³.

Show that the sum of the binomial coefficients of (a + b)ⁿ is 2ⁿ. (3 marks)

By the binomial theorem, (a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + … + ⁿCₙbⁿ. Put a = 1 and b = 1. The left side becomes 2ⁿ. The right side becomes ⁿC₀ + ⁿC₁ + … + ⁿCₙ. So the sum of the coefficients is 2ⁿ.

Prove by induction that 2 + 4 + 6 + … + 2n = n(n + 1). (5 marks)

For n = 1, the left side is 2 and the right side is 1 × 2 = 2, so it is true. Assume it is true for n = k: 2 + 4 + … + 2k = k(k + 1). Add the next term 2(k + 1) to both sides. The left side becomes 2 + 4 + … + 2k + 2(k + 1). The right side becomes k(k + 1) + 2(k + 1) = (k + 1)(k + 2), which is the formula with n = k + 1. So the statement is true for n = k + 1. By induction it is true for all natural n.

Explain how to find the term independent of x in the expansion of (x² + 1/x)ⁿ. (5 marks)

Write the general term Tᵣ₊₁ = ⁿCᵣ (x²)ⁿ⁻ʳ (1/x)ʳ = ⁿCᵣ x²ⁿ⁻²ʳ⁻ʳ = ⁿCᵣ x²ⁿ⁻³ʳ. The term is independent of x when the power of x is 0, so 2n − 3r = 0 and r = 2n/3. This is valid only when r is a whole number between 0 and n. Substitute this r into ⁿCᵣ to get the term. If r is not a whole number, there is no such term.

Numericals with worked steps

Expand (x + 2)⁴.

The coefficients for n = 4 are 1, 4, 6, 4, 1.
Terms: x⁴, 4x³(2), 6x²(2²), 4x(2³), 2⁴.
= x⁴ + 8x³ + 6x²(4) + 4x(8) + 16.
= x⁴ + 8x³ + 24x² + 32x + 16.

Answer: x⁴ + 8x³ + 24x² + 32x + 16.

Find the 6th term of (2x + 1)⁸.

T₆ = Tᵣ₊₁ with r = 5.
T₆ = ⁸C₅ (2x)³ (1)⁵.
⁸C₅ = ⁸C₃ = (8 × 7 × 6)/(3 × 2 × 1) = 56.
(2x)³ = 8x³, so T₆ = 56 × 8x³ = 448x³.

Answer: T₆ = 448x³.

Find the term independent of x in (x² + 1/x)⁹.

Tᵣ₊₁ = ⁹Cᵣ (x²)⁹⁻ʳ (1/x)ʳ = ⁹Cᵣ x¹⁸⁻³ʳ.
For the term independent of x, 18 − 3r = 0, so r = 6.
⁹C₆ = ⁹C₃ = (9 × 8 × 7)/(3 × 2 × 1) = 84.

Answer: The term independent of x is 84 (the 7th term).

Find (1.02)⁴ correct to 4 decimal places using the binomial theorem.

Write (1 + 0.02)⁴ = 1 + 4(0.02) + 6(0.02)² + 4(0.02)³ + (0.02)⁴.
4(0.02) = 0.08 and 6(0.0004) = 0.0024.
4(0.000008) = 0.000032, and (0.02)⁴ = 0.00000016.
Sum: 1 + 0.08 + 0.0024 + 0.000032 + 0.00000016 = 1.08243216.
To 4 decimal places this is 1.0824.

Answer: (1.02)⁴ ≈ 1.0824.

Common mistakes

Skipping the basis step, or proving only the induction step. Both are needed.
In the induction step, using P(k + 1) as a starting point instead of proving it from P(k).
Using r instead of r + 1 for the term number, such as taking T₆ with r = 6 instead of r = 5.
Forgetting the alternating signs in the expansion of (a − b)ⁿ.
Losing the coefficient of the term inside a bracket, such as writing (2x)³ as 2x³ instead of 8x³.

Practise this chapter

Practice questions and timed test Printable worksheet