7.4 Q-5 | Mathematical Induction And Binomial Theorem | Mathematics 11

Exercise 7.4 — Question 5

This question applies the Binomial Theorem to practical problems: finding remainders, last digits, approximate values, and simplifying expanded expressions.

Key Techniques

1. Expanding and Simplifying (SLO M-11-A-37)

When two binomial expansions are added or subtracted, many terms cancel:

Difference $(a + b)^{n} - (a - b)^{n}$ : even-powered terms cancel, odd-powered terms double.
Sum $(a + b)^{n} + (a - b)^{n}$ : odd-powered terms cancel, even-powered terms double.

Example: Simplify $(x + 2)^{5} - (x - 2)^{5}$

$(x + 2)^{5} - (x - 2)^{5} = 2 [(1 5) x^{4} (2) + (3 5) x^{2} (8) + (5 5) (32)]$ $= 2 [10 x^{4} + 80 x^{2} + 32] = 20 x^{4} + 160 x^{2} + 64$

2. Finding Remainders (SLO M-11-A-39)

Write the base as $1 + (multiple of divisor)$ , expand, and identify the remainder.

Example: Find the remainder when $7^{100}$ is divided by $6$ .

$7^{100} = (6 + 1)^{100} = \sum_{r = 0}^{100} (r 100) 6^{r} (1)^{100 - r}$

All terms with $r \geq 1$ are divisible by $6$ . The only term not divisible by $6$ is the $r = 0$ term: $(0 100) 6^{0} = 1$

$∴ Remainder = 1$

3. Finding the Last Digit (SLO M-11-A-40)

The last digit of $N^{n}$ depends only on the last digit of $N$ . Use the Binomial Theorem by writing $N = (multiple of 10) + d$ where $d$ is the units digit.

Example: Find the last digit of $3^{100}$ .

$3^{100} = (3^{4})^{25} = 8 1^{25} = (80 + 1)^{25}$

Expanding: all terms except the last contain a factor of $80$ (divisible by $10$ ), so: $8 1^{25} \equiv 1^{25} = 1 (mod 10)$

$∴ Last digit of 3^{100} = 1$

4. Approximate Values (SLO M-11-A-38)

For $∣ x ∣ ≪ 1$ , higher powers of $x$ are negligible. Use the first few terms of $(1 + x)^{n}$ :

$(1 + x)^{n} \approx 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \dots$

Example: Approximate $(1.02)^{10}$ .

$(1.02)^{10} = (1 + 0.02)^{10} \approx 1 + 10 (0.02) + \frac{10 \cdot 9}{2} (0.02)^{2}$ $\approx 1 + 0.2 + 45 (0.0004) = 1 + 0.2 + 0.018 = 1.218$

Summary Table

Technique	Key Idea
Expand & Simplify	Cancel even/odd terms in sum/difference
Remainder	Write base as $1 + k \cdot d$ ; remainder from constant term
Last Digit	Reduce mod 10; use periodicity of units digits
Approximation	Keep first 2–3 terms when $

Exercise 7.4 — Question 5

This question applies the Binomial Theorem to practical problems: finding remainders, last digits, approximate values, and simplifying expanded expressions.

Key Techniques

1. Expanding and Simplifying (SLO M-11-A-37)

When two binomial expansions are added or subtracted, many terms cancel:

Difference $(a + b)^{n} - (a - b)^{n}$ : even-powered terms cancel, odd-powered terms double.
Sum $(a + b)^{n} + (a - b)^{n}$ : odd-powered terms cancel, even-powered terms double.

Example: Simplify $(x + 2)^{5} - (x - 2)^{5}$

$(x + 2)^{5} - (x - 2)^{5} = 2 [(1 5) x^{4} (2) + (3 5) x^{2} (8) + (5 5) (32)]$ $= 2 [10 x^{4} + 80 x^{2} + 32] = 20 x^{4} + 160 x^{2} + 64$

2. Finding Remainders (SLO M-11-A-39)

Write the base as $1 + (multiple of divisor)$ , expand, and identify the remainder.

Example: Find the remainder when $7^{100}$ is divided by $6$ .

$7^{100} = (6 + 1)^{100} = \sum_{r = 0}^{100} (r 100) 6^{r} (1)^{100 - r}$

All terms with $r \geq 1$ are divisible by $6$ . The only term not divisible by $6$ is the $r = 0$ term: $(0 100) 6^{0} = 1$

$∴ Remainder = 1$

3. Finding the Last Digit (SLO M-11-A-40)

The last digit of $N^{n}$ depends only on the last digit of $N$ . Use the Binomial Theorem by writing $N = (multiple of 10) + d$ where $d$ is the units digit.

Example: Find the last digit of $3^{100}$ .

$3^{100} = (3^{4})^{25} = 8 1^{25} = (80 + 1)^{25}$

Expanding: all terms except the last contain a factor of $80$ (divisible by $10$ ), so: $8 1^{25} \equiv 1^{25} = 1 (mod 10)$

$∴ Last digit of 3^{100} = 1$

4. Approximate Values (SLO M-11-A-38)

For $∣ x ∣ ≪ 1$ , higher powers of $x$ are negligible. Use the first few terms of $(1 + x)^{n}$ :

$(1 + x)^{n} \approx 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \dots$

Example: Approximate $(1.02)^{10}$ .

$(1.02)^{10} = (1 + 0.02)^{10} \approx 1 + 10 (0.02) + \frac{10 \cdot 9}{2} (0.02)^{2}$ $\approx 1 + 0.2 + 45 (0.0004) = 1 + 0.2 + 0.018 = 1.218$

Summary Table

Technique	Key Idea
Expand & Simplify	Cancel even/odd terms in sum/difference
Remainder	Write base as $1 + k \cdot d$ ; remainder from constant term
Last Digit	Reduce mod 10; use periodicity of units digits
Approximation	Keep first 2–3 terms when $