7.2 Q-9 | Mathematical Induction And Binomial Theorem | Mathematics 11

Exercise 7.2 — Question 9

This question applies the Binomial Theorem to practical problems involving:

Finding remainders when large powers are divided by a number
Finding the last digit of a large power
Testing divisibility
Finding approximate values

Key Formula Reminder

The general term of the binomial expansion $(a + b)^{n}$ is:

$T_{r + 1} = (r n) a^{n - r} b^{r}, r = 0, 1, 2, \dots, n$

Technique 1 — Finding Remainders

To find the remainder when a large power is divided by a number, rewrite the base so that one part is a multiple of the divisor.

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

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

Expanding: $= (0 100) 6^{100} + (1 100) 6^{99} + \dots + (99 100) 6^{1} + (100 100) 6^{0}$

Every term except the last contains a factor of $6$ , so: $7^{100} = 6 k + 1 for some integer k$

$Remainder = 1$

Technique 2 — Finding the Last Digit

The last digit of a number equals its remainder when divided by 10.

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

Solution: $3^{100} = (10 - 7)^{100}$

Alternatively, use the pattern of powers of 3: $3^{1} = 3, 3^{2} = 9, 3^{3} = 27, 3^{4} = 81, 3^{5} = 243, \dots$

The last digits cycle with period 4: $3, 9, 7, 1, 3, 9, 7, 1, \dots$

Since $100 = 4 \times 25$ , we have $3^{100} = (3^{4})^{25} = 8 1^{25}$ .

$Last digit of 3^{100} = 1$

Technique 3 — Approximate Values

For expressions like $(1 + x)^{n}$ where $x$ is small, take only the first few terms of the expansion.

Example: Find the approximate value of $(1.02)^{10}$ .

Solution: $(1.02)^{10} = (1 + 0.02)^{10}$

Using the first three terms: $\approx 1 + (1 10) (0.02) + (2 10) (0.02)^{2}$ $= 1 + 10 (0.02) + 45 (0.0004)$ $= 1 + 0.2 + 0.018$ $\approx 1.218$

Middle Terms

For $(a + b)^{n}$ with $n + 1$ total terms:

If $n$ is even: one middle term $T_{\frac{n}{2} + 1}$
If $n$ is odd: two middle terms $T_{\frac{n + 1}{2}}$ and $T_{\frac{n + 3}{2}}$

Exercise 7.2 — Question 9

This question applies the Binomial Theorem to practical problems involving:

Finding remainders when large powers are divided by a number
Finding the last digit of a large power
Testing divisibility
Finding approximate values

Key Formula Reminder

The general term of the binomial expansion $(a + b)^{n}$ is:

$T_{r + 1} = (r n) a^{n - r} b^{r}, r = 0, 1, 2, \dots, n$

Technique 1 — Finding Remainders

To find the remainder when a large power is divided by a number, rewrite the base so that one part is a multiple of the divisor.

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

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

Expanding: $= (0 100) 6^{100} + (1 100) 6^{99} + \dots + (99 100) 6^{1} + (100 100) 6^{0}$

Every term except the last contains a factor of $6$ , so: $7^{100} = 6 k + 1 for some integer k$

$Remainder = 1$

Technique 2 — Finding the Last Digit

The last digit of a number equals its remainder when divided by 10.

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

Solution: $3^{100} = (10 - 7)^{100}$

Alternatively, use the pattern of powers of 3: $3^{1} = 3, 3^{2} = 9, 3^{3} = 27, 3^{4} = 81, 3^{5} = 243, \dots$

The last digits cycle with period 4: $3, 9, 7, 1, 3, 9, 7, 1, \dots$

Since $100 = 4 \times 25$ , we have $3^{100} = (3^{4})^{25} = 8 1^{25}$ .

$Last digit of 3^{100} = 1$

Technique 3 — Approximate Values

For expressions like $(1 + x)^{n}$ where $x$ is small, take only the first few terms of the expansion.

Example: Find the approximate value of $(1.02)^{10}$ .

Solution: $(1.02)^{10} = (1 + 0.02)^{10}$

Using the first three terms: $\approx 1 + (1 10) (0.02) + (2 10) (0.02)^{2}$ $= 1 + 10 (0.02) + 45 (0.0004)$ $= 1 + 0.2 + 0.018$ $\approx 1.218$

Middle Terms

For $(a + b)^{n}$ with $n + 1$ total terms:

If $n$ is even: one middle term $T_{\frac{n}{2} + 1}$
If $n$ is odd: two middle terms $T_{\frac{n + 1}{2}}$ and $T_{\frac{n + 3}{2}}$