7.3 Binomial Series and Convergence | Mathematical Induction And Binomial Theorem | Mathematics 11

Overview

Exercise 7.3 extends the Binomial Theorem beyond positive integer exponents. When the exponent $n$ is a fraction or a negative integer, the expansion becomes an infinite series called the Binomial Series, and convergence must be checked.

The Binomial Series

For any real number $n$ (including fractions and negative integers):

$(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \frac{n ( n - 1 ) ( n - 2 )}{3 !} x^{3} + \dots$

Convergence condition: The series is valid (converges) only when $∣ x ∣ < 1$ .

General Term

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

Convergence Condition

For the binomial series to converge, the variable term inside the bracket must satisfy $∣ x ∣ < 1$ .

Expanding $(a + x)^{n}$

When the expression is not in the form $(1 + x)^{n}$ , factor out the constant first:

$(a + x)^{n} = a^{n} (1 + \frac{x}{a})^{n}, valid when \frac{x}{a} < 1$

Example: Expand $(2 + 3 x)^{- 1}$ for $∣ x ∣ < \frac{2}{3}$ :

$(2 + 3 x)^{- 1} = 2^{- 1} (1 + \frac{3 x}{2})^{- 1} = \frac{1}{2} [1 - \frac{3 x}{2} + \frac{9 x ^{2}}{4} - \dots]$

Approximations Using the Binomial Series

First-Order Approximation (small $x$ )

When $x$ is so small that $x^{2}$ and higher powers are negligible:

$(1 + x)^{n} \approx 1 + n x$

Second-Order Approximation

When $x^{3}$ and higher powers are negligible:

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

Approximation When $x$ is Large

When $x$ is very large, factor out the highest power of $x$ :

$(x + a)^{n} = x^{n} (1 + \frac{a}{x})^{n}, valid when \frac{a}{x} < 1$

Sign Pattern in Expansions with Negative Exponents

For negative exponents, successive terms alternate in sign:

$(1 + x)^{- 3} = 1 - 3 x + 6 x^{2} - 10 x^{3} + \dots$

Finding Remainders Using the Binomial Theorem

To find the remainder when a large power is divided by a number, rewrite the base as $(1 + k)^{n}$ or $(m + 1)^{n}$ where $m$ is the divisor.

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

$7^{100} = (6 + 1)^{100} = (0 100) 6^{0} + (1 100) 6^{1} + (2 100) 6^{2} + \dots$

All terms with $6^{1}$ or higher are divisible by $6$ . The only term not divisible by $6$ is $(0 100) \cdot 1 = 1$ .

$∴ 7^{100} \equiv 1 (mod 6)$

Finding the Last Digit of a Large Power

To find the units digit of $N^{k}$ , find $N^{k} (mod 10)$ . Use the Binomial Theorem by writing $N = (10 + d)$ where $d$ is the units digit of $N$ .

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

The units digits of powers of $3$ cycle: $3, 9, 7, 1, 3, 9, 7, 1, \dots$ (period 4).

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

The last digit of $8 1^{25}$ is the last digit of $1^{25} = 1$ .

$∴ Last digit of 3^{100} = 1$

Summary Table

Situation	Action
Expression is $(a + x)^{n}$	Factor out $a^{n}$ first
$x$ is very small	Use $(1 + x)^{n} \approx 1 + n x$
$x$ is very large	Factor out highest power of $x$
Finding remainder $N^{k} \div m$	Write $N = (m \pm 1)^{k}$ and expand
Finding last digit	Find $N^{k} (mod 10)$ using cycle of units digits

Overview

The Binomial Series

For any real number $n$ (including fractions and negative integers):

$(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \frac{n ( n - 1 ) ( n - 2 )}{3 !} x^{3} + \dots$

Convergence condition: The series is valid (converges) only when $∣ x ∣ < 1$ .

General Term

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

Convergence Condition

For the binomial series to converge, the variable term inside the bracket must satisfy $∣ x ∣ < 1$ .

Expanding $(a + x)^{n}$

When the expression is not in the form $(1 + x)^{n}$ , factor out the constant first:

$(a + x)^{n} = a^{n} (1 + \frac{x}{a})^{n}, valid when \frac{x}{a} < 1$

Example: Expand $(2 + 3 x)^{- 1}$ for $∣ x ∣ < \frac{2}{3}$ :

$(2 + 3 x)^{- 1} = 2^{- 1} (1 + \frac{3 x}{2})^{- 1} = \frac{1}{2} [1 - \frac{3 x}{2} + \frac{9 x ^{2}}{4} - \dots]$

Approximations Using the Binomial Series

First-Order Approximation (small $x$ )

When $x$ is so small that $x^{2}$ and higher powers are negligible:

$(1 + x)^{n} \approx 1 + n x$

Second-Order Approximation

When $x^{3}$ and higher powers are negligible:

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

Approximation When $x$ is Large

When $x$ is very large, factor out the highest power of $x$ :

$(x + a)^{n} = x^{n} (1 + \frac{a}{x})^{n}, valid when \frac{a}{x} < 1$

Sign Pattern in Expansions with Negative Exponents

For negative exponents, successive terms alternate in sign:

$(1 + x)^{- 3} = 1 - 3 x + 6 x^{2} - 10 x^{3} + \dots$

Finding Remainders Using the Binomial Theorem

To find the remainder when a large power is divided by a number, rewrite the base as $(1 + k)^{n}$ or $(m + 1)^{n}$ where $m$ is the divisor.

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

$7^{100} = (6 + 1)^{100} = (0 100) 6^{0} + (1 100) 6^{1} + (2 100) 6^{2} + \dots$

All terms with $6^{1}$ or higher are divisible by $6$ . The only term not divisible by $6$ is $(0 100) \cdot 1 = 1$ .

$∴ 7^{100} \equiv 1 (mod 6)$

Finding the Last Digit of a Large Power

To find the units digit of $N^{k}$ , find $N^{k} (mod 10)$ . Use the Binomial Theorem by writing $N = (10 + d)$ where $d$ is the units digit of $N$ .

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

The units digits of powers of $3$ cycle: $3, 9, 7, 1, 3, 9, 7, 1, \dots$ (period 4).

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

The last digit of $8 1^{25}$ is the last digit of $1^{25} = 1$ .

$∴ Last digit of 3^{100} = 1$

Summary Table

Situation	Action
Expression is $(a + x)^{n}$	Factor out $a^{n}$ first
$x$ is very small	Use $(1 + x)^{n} \approx 1 + n x$
$x$ is very large	Factor out highest power of $x$
Finding remainder $N^{k} \div m$	Write $N = (m \pm 1)^{k}$ and expand
Finding last digit	Find $N^{k} (mod 10)$ using cycle of units digits