7.2 Q-18 | Mathematical Induction And Binomial Theorem | Mathematics 11

Exercise 7.2 — Question 18

Topic: Applications of the Binomial Theorem

This question applies the Binomial Theorem to practical problems such as finding middle terms, remainders, and last digits of large powers.

Key Formula: General Term

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

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

The total number of terms in the expansion is $n + 1$ .

Middle Term(s)

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

Example: Find the middle term of $(x + y)^{8}$ .

Here $n = 8$ (even), so the middle term is $T_{\frac{8}{2} + 1} = T_{5}$ .

$T_{5} = (4 8) x^{8 - 4} y^{4} = 70 x^{4} y^{4}$

Finding the Remainder Using the Binomial Theorem (SLO M-11-A-39)

To find the remainder when a large power is divided by a number, rewrite the base as a sum involving that number.

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

Write $7 = 6 + 1$ , so:

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

$= (0 100) + (1 100) \cdot 6 + (2 100) \cdot 6^{2} + \dots$

Every term except the first contains a factor of $6$ , so:

$7^{100} = 1 + 6 [(1 100) + (2 100) \cdot 6 + \dots]$

Therefore $7^{100} \equiv 1 (mod 6)$ . The remainder is 1.

Finding the Last Digit Using the Binomial Theorem (SLO M-11-A-40)

To find the last (units) digit of a large power, find the remainder when divided by $10$ .

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

Note that $3^{4} = 81$ , so $3^{100} = (3^{4})^{25} = 8 1^{25}$ .

Write $81 = 80 + 1$ :

$8 1^{25} = (80 + 1)^{25} = \sum_{r = 0}^{25} (r 25) 8 0^{r}$

$= 1 + 25 \cdot 80 + (2 25) \cdot 8 0^{2} + \dots$

Every term except the first is divisible by $10$ (since $80$ is divisible by $10$ ), so:

$3^{100} \equiv 1 (mod 10)$

The last digit of $3^{100}$ is 1.

Testing Divisibility Using the Binomial Theorem

To show that an expression is divisible by a number $k$ , rewrite the base so that all terms in the expansion are multiples of $k$ except possibly the constant term, then check the constant.

Example: Show that $5^{n} - 1$ is divisible by $4$ for all positive integers $n$ .

Write $5 = 4 + 1$ :

$5^{n} = (4 + 1)^{n} = \sum_{r = 0}^{n} (r n) 4^{r} = 1 + n \cdot 4 + (2 n) \cdot 16 + \dots$

$5^{n} - 1 = 4 [n + (2 n) \cdot 4 + \dots]$

This is clearly divisible by $4$ . $■$

Exercise 7.2 — Question 18

Topic: Applications of the Binomial Theorem

This question applies the Binomial Theorem to practical problems such as finding middle terms, remainders, and last digits of large powers.

Key Formula: General Term

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

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

The total number of terms in the expansion is $n + 1$ .

Middle Term(s)

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

Example: Find the middle term of $(x + y)^{8}$ .

Here $n = 8$ (even), so the middle term is $T_{\frac{8}{2} + 1} = T_{5}$ .

$T_{5} = (4 8) x^{8 - 4} y^{4} = 70 x^{4} y^{4}$

Finding the Remainder Using the Binomial Theorem (SLO M-11-A-39)

To find the remainder when a large power is divided by a number, rewrite the base as a sum involving that number.

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

Write $7 = 6 + 1$ , so:

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

$= (0 100) + (1 100) \cdot 6 + (2 100) \cdot 6^{2} + \dots$

Every term except the first contains a factor of $6$ , so:

$7^{100} = 1 + 6 [(1 100) + (2 100) \cdot 6 + \dots]$

Therefore $7^{100} \equiv 1 (mod 6)$ . The remainder is 1.

Finding the Last Digit Using the Binomial Theorem (SLO M-11-A-40)

To find the last (units) digit of a large power, find the remainder when divided by $10$ .

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

Note that $3^{4} = 81$ , so $3^{100} = (3^{4})^{25} = 8 1^{25}$ .

Write $81 = 80 + 1$ :

$8 1^{25} = (80 + 1)^{25} = \sum_{r = 0}^{25} (r 25) 8 0^{r}$

$= 1 + 25 \cdot 80 + (2 25) \cdot 8 0^{2} + \dots$

Every term except the first is divisible by $10$ (since $80$ is divisible by $10$ ), so:

$3^{100} \equiv 1 (mod 10)$

The last digit of $3^{100}$ is 1.

Testing Divisibility Using the Binomial Theorem

To show that an expression is divisible by a number $k$ , rewrite the base so that all terms in the expansion are multiples of $k$ except possibly the constant term, then check the constant.

Example: Show that $5^{n} - 1$ is divisible by $4$ for all positive integers $n$ .

Write $5 = 4 + 1$ :

$5^{n} = (4 + 1)^{n} = \sum_{r = 0}^{n} (r n) 4^{r} = 1 + n \cdot 4 + (2 n) \cdot 16 + \dots$

$5^{n} - 1 = 4 [n + (2 n) \cdot 4 + \dots]$

This is clearly divisible by $4$ . $■$