7.2 Binomial Theorem and Pascal's Triangle | Mathematical Induction And Binomial Theorem | Mathematics 11

The Binomial Theorem

The Binomial Theorem gives a formula for expanding any power of a binomial $(a + b)^{n}$ where $n$ is a positive integer (natural number):

$(a + b)^{n} = \sum_{r = 0}^{n} (r n) a^{n - r} b^{r}$

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

The theorem is restricted to natural numbers $n \in N$ in this form (SLO M-11-A-35).

Pascal's Triangle and Binomial Coefficients

The binomial coefficients $(r n) = \frac{n !}{r ! ( n - r )!}$ can be read directly from Pascal's Triangle:

n=0:          1
n=1:        1   1
n=2:      1   2   1
n=3:    1   3   3   1
n=4:  1   4   6   4   1
n=5: 1  5  10  10   5  1

Each entry is the sum of the two entries directly above it. The $(n + 1)^{t h}$ row gives the coefficients of $(a + b)^{n}$ .

General Term (The $r^{t h}$ Term)

The general term (also called the $(r + 1)^{t h}$ term) is:

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

Note: $r$ is a zero-based index. The first term has $r = 0$ , the second has $r = 1$ , etc.

Expansion of $(a - b)^{n}$

Replace $b$ with $- b$ :

$T_{r + 1} = (r n) a^{n - r} (- b)^{r} = (- 1)^{r} (r n) a^{n - r} b^{r}$

Terms with odd $r$ are negative; terms with even $r$ are positive.

Middle Terms

The expansion of $(a + b)^{n}$ has $n + 1$ terms.

$n$	Number of middle terms	Position(s)
Even	1	$T_{\frac{n}{2} + 1}$
Odd	2	$T_{\frac{n + 1}{2}}$ and $T_{\frac{n + 3}{2}}$

The $r^{t h}$ term from the end of $(a + b)^{n}$ is the same as the $r^{t h}$ term from the beginning of $(b + a)^{n}$ , which equals the $(n - r + 2)^{t h}$ term from the start of $(a + b)^{n}$ .

Term Independent of $x$

To find the constant term (term independent of $x$ ):

Write the general term $T_{r + 1}$ .
Collect all powers of $x$ into $x^{f (r)}$ .
Solve $f (r) = 0$ for $r$ .
Substitute back to find the term.

Example: Find the term independent of $x$ in $(x^{2} + \frac{1}{x})^{9}$ .

$T_{r + 1} = (r 9) (x^{2})^{9 - r} (\frac{1}{x})^{r} = (r 9) x^{18 - 2 r - r} = (r 9) x^{18 - 3 r}$

Set $18 - 3 r = 0 \Rightarrow r = 6$ .

$T_{7} = (6 9) = 84$

Coefficient of a Specific Power

When the expansion is multiplied by another polynomial, find the coefficient of $x^{k}$ by:

Expanding each factor.
Identifying all pairs of terms whose powers of $x$ sum to $k$ .
Summing the products of their coefficients.

Expanding Trinomials

To expand $(a + b + c)^{n}$ , group two terms:

$[(a + b) + c]^{n} = \sum_{r = 0}^{n} (r n) (a + b)^{n - r} c^{r}$

Then expand each $(a + b)^{n - r}$ using the Binomial Theorem again.

Important Binomial Coefficient Identities

Sum of All Coefficients

Substitute $a = 1, b = 1$ into $(a + b)^{n}$ :

$\sum_{r = 0}^{n} (r n) = 2^{n}$

Sum of Even- and Odd-Indexed Coefficients

Substitute $a = 1, b = - 1$ :

$(0 n) - (1 n) + (2 n) - \dots = 0 ⟹ S_{e} = S_{o}$

Since $S_{e} + S_{o} = 2^{n}$ :

$S_{e} = S_{o} = 2^{n - 1}$

Approximate Values Using the Binomial Theorem

For small $x$ , higher powers become negligible. Write the number as $(1 \pm x)^{n}$ and keep the first few terms.

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

$(1 + 0.02)^{10} \approx 1 + 10 (0.02) + (2 10) (0.02)^{2} + (3 10) (0.02)^{3}$ $= 1 + 0.2 + 45 (0.0004) + 120 (0.000008) = 1 + 0.2 + 0.018 + 0.00096 \approx 1.219$

Remainder Using the Binomial Theorem

To find the remainder when $N^{m}$ is divided by $d$ , write $N = (d \pm k)$ so that $N^{m} = (d \pm k)^{m}$ . Expand; all terms with $d$ as a factor vanish modulo $d$ , leaving only the constant term.

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

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

All terms with $r \geq 1$ are divisible by 6. The $r = 0$ term is $1$ .

$∴ 7^{100} \equiv 1 (mod 6)$

Last Digit and Divisibility

To find the last digit of $N^{m}$ , find $N^{m} (mod 10)$ . Express $N^{m}$ as $(10 k \pm c)^{m}$ and expand.

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

$3^{100} = (3^{2})^{50} = 9^{50} = (10 - 1)^{50} = \sum_{r = 0}^{50} (r 50) 1 0^{r} (- 1)^{50 - r}$

All terms with $r \geq 1$ are divisible by 10. The $r = 0$ term is $(- 1)^{50} = 1$ .

$∴ Last digit of 3^{100} = 1$

Real-World Applications

The Binomial Theorem and Pascal's Triangle appear in many real-world contexts (SLO M-11-A-41):

Probability — The binomial distribution uses $(r n) p^{r} (1 - p)^{n - r}$ to model outcomes of repeated trials.
Economic Forecasting — Compound growth $(1 + r)^{n}$ is approximated using the first few terms of the Binomial expansion.
Pascal's Triangle Patterns — Rows encode combinatorial identities, Fibonacci numbers appear as diagonal sums, and powers of 11 are read directly from rows.
Domino Effects and Rankings — Counting arrangements and cascading outcomes use binomial coefficients.
Variable Substitution — Complex algebraic expressions are simplified by substituting grouped terms as a single binomial.
Puzzles — Problems involving large powers, last digits, and divisibility are solved elegantly using the Binomial Theorem.

The Binomial Theorem

The Binomial Theorem gives a formula for expanding any power of a binomial $(a + b)^{n}$ where $n$ is a positive integer (natural number):

$(a + b)^{n} = \sum_{r = 0}^{n} (r n) a^{n - r} b^{r}$

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

The theorem is restricted to natural numbers $n \in N$ in this form (SLO M-11-A-35).

Pascal's Triangle and Binomial Coefficients

The binomial coefficients $(r n) = \frac{n !}{r ! ( n - r )!}$ can be read directly from Pascal's Triangle:

n=0:          1
n=1:        1   1
n=2:      1   2   1
n=3:    1   3   3   1
n=4:  1   4   6   4   1
n=5: 1  5  10  10   5  1

Each entry is the sum of the two entries directly above it. The $(n + 1)^{t h}$ row gives the coefficients of $(a + b)^{n}$ .

General Term (The $r^{t h}$ Term)

The general term (also called the $(r + 1)^{t h}$ term) is:

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

Note: $r$ is a zero-based index. The first term has $r = 0$ , the second has $r = 1$ , etc.

Expansion of $(a - b)^{n}$

Replace $b$ with $- b$ :

$T_{r + 1} = (r n) a^{n - r} (- b)^{r} = (- 1)^{r} (r n) a^{n - r} b^{r}$

Terms with odd $r$ are negative; terms with even $r$ are positive.

Middle Terms

The expansion of $(a + b)^{n}$ has $n + 1$ terms.

$n$	Number of middle terms	Position(s)
Even	1	$T_{\frac{n}{2} + 1}$
Odd	2	$T_{\frac{n + 1}{2}}$ and $T_{\frac{n + 3}{2}}$

Term from the End

Term Independent of $x$

To find the constant term (term independent of $x$ ):

Write the general term $T_{r + 1}$ .
Collect all powers of $x$ into $x^{f (r)}$ .
Solve $f (r) = 0$ for $r$ .
Substitute back to find the term.

Example: Find the term independent of $x$ in $(x^{2} + \frac{1}{x})^{9}$ .

$T_{r + 1} = (r 9) (x^{2})^{9 - r} (\frac{1}{x})^{r} = (r 9) x^{18 - 2 r - r} = (r 9) x^{18 - 3 r}$

Set $18 - 3 r = 0 \Rightarrow r = 6$ .

$T_{7} = (6 9) = 84$

Coefficient of a Specific Power

When the expansion is multiplied by another polynomial, find the coefficient of $x^{k}$ by:

Expanding each factor.
Identifying all pairs of terms whose powers of $x$ sum to $k$ .
Summing the products of their coefficients.

Expanding Trinomials

To expand $(a + b + c)^{n}$ , group two terms:

$[(a + b) + c]^{n} = \sum_{r = 0}^{n} (r n) (a + b)^{n - r} c^{r}$

Then expand each $(a + b)^{n - r}$ using the Binomial Theorem again.

Important Binomial Coefficient Identities

Sum of All Coefficients

Substitute $a = 1, b = 1$ into $(a + b)^{n}$ :

$\sum_{r = 0}^{n} (r n) = 2^{n}$

Sum of Even- and Odd-Indexed Coefficients

Substitute $a = 1, b = - 1$ :

$(0 n) - (1 n) + (2 n) - \dots = 0 ⟹ S_{e} = S_{o}$

Since $S_{e} + S_{o} = 2^{n}$ :

$S_{e} = S_{o} = 2^{n - 1}$

Approximate Values Using the Binomial Theorem

For small $x$ , higher powers become negligible. Write the number as $(1 \pm x)^{n}$ and keep the first few terms.

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

$(1 + 0.02)^{10} \approx 1 + 10 (0.02) + (2 10) (0.02)^{2} + (3 10) (0.02)^{3}$ $= 1 + 0.2 + 45 (0.0004) + 120 (0.000008) = 1 + 0.2 + 0.018 + 0.00096 \approx 1.219$

Remainder Using the Binomial Theorem

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

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

All terms with $r \geq 1$ are divisible by 6. The $r = 0$ term is $1$ .

$∴ 7^{100} \equiv 1 (mod 6)$

Last Digit and Divisibility

To find the last digit of $N^{m}$ , find $N^{m} (mod 10)$ . Express $N^{m}$ as $(10 k \pm c)^{m}$ and expand.

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

$3^{100} = (3^{2})^{50} = 9^{50} = (10 - 1)^{50} = \sum_{r = 0}^{50} (r 50) 1 0^{r} (- 1)^{50 - r}$

All terms with $r \geq 1$ are divisible by 10. The $r = 0$ term is $(- 1)^{50} = 1$ .

$∴ Last digit of 3^{100} = 1$

Real-World Applications

The Binomial Theorem and Pascal's Triangle appear in many real-world contexts (SLO M-11-A-41):

Probability — The binomial distribution uses $(r n) p^{r} (1 - p)^{n - r}$ to model outcomes of repeated trials.
Economic Forecasting — Compound growth $(1 + r)^{n}$ is approximated using the first few terms of the Binomial expansion.
Pascal's Triangle Patterns — Rows encode combinatorial identities, Fibonacci numbers appear as diagonal sums, and powers of 11 are read directly from rows.
Domino Effects and Rankings — Counting arrangements and cascading outcomes use binomial coefficients.
Variable Substitution — Complex algebraic expressions are simplified by substituting grouped terms as a single binomial.
Puzzles — Problems involving large powers, last digits, and divisibility are solved elegantly using the Binomial Theorem.