Binomial Theorem notes: CBSE Class 11 Mathematics

1. Pascal's triangle

Write $1$ at the top. Every row begins and ends with $1$ and each inner entry is the sum of the two entries above it. Row $n$ (the top row is row $0$ ) lists the numbers $\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n}$ , which are the coefficients of $(a + b)^{n}$ .

The rows are symmetric, and row $n$ adds up to $2^{n}$ . Row $7$ is $1, 7, 21, 35, 35, 21, 7, 1$ .

The triangle was known long before Pascal: Pingala in India (about the second century before the common era), Persian and Chinese mathematicians (Yang Hui, about the thirteenth century) and later Pascal (about the 1650s). Pascal's triangle is the usual name in Europe.

Each inner number is the sum of the two above: $15 + 20 = 35$.

The coefficients form a pattern.
Power $n$	Expansion of $(a + b)^{n}$ by multiplying the brackets out
$0$	$1$
$1$	$a + b$
$2$	$a^{2} + 2ab + b^{2}$
$3$	$a^{3} + 3a^{2}b + 3ab^{2} + b^{3}$
$4$	$a^{4} + 4a^{3}b + 6a^{2}b^{2} + 4ab^{3} + b^{4}$
$5$	$a^{5} + 5a^{4}b + 10a^{3}b^{2} + 10a^{2}b^{3} + 5ab^{4} + b^{5}$

The same triangle was found in many places.
Place and time (about)	Who	What they did
India, about the second century before the common era	Pingala	Counted patterns of short and long syllables in poetry with these numbers; later writers such as Halayudha (tenth century) set them out as the "staircase of Mount Meru"
Persia, about the eleventh and twelfth centuries	Al-Karaji and Omar Khayyam	Used the number pattern to expand powers of a sum and to take roots
China, about the thirteenth century	Yang Hui	Published the triangle, still called Yang Hui's triangle in China
France, about the 1650s	Blaise Pascal	Studied the triangle closely and used it in counting and chance, so Europe names it after him
England, about the 1660s	Isaac Newton	Extended the expansion beyond whole-number powers (not needed in this chapter)

Formulas

Pascal's rule
$\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$
Symmetry
$\binom{n}{r} = \binom{n}{n-r}$

Key terms

Pascal's triangle: A triangle of numbers in which each inner entry is the sum of the two above it; row $n$ gives the coefficients of $(a+b)^{n}$ .
Binomial coefficient: $\binom{n}{r} = {}^{n}C_{r} = \dfrac{n!}{r!\,(n-r)!}$ , the number of ways to choose $r$ of $n$ objects.

2. The binomial theorem and its proof

For every positive integer $n$ ,

(a + b)^{n} = \binom{n}{0}a^{n} + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^{2} + \cdots + \binom{n}{n}b^{n} = \sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^{r}

Proof by counting. The product of the $n$ brackets $(a + b)(a + b)\cdots(a + b)$ is formed by choosing $a$ or $b$ from every bracket and multiplying. If $b$ is chosen from exactly $r$ brackets, the term is $a^{n-r}b^{r}$ , and there are $\binom{n}{r}$ ways to pick those brackets. Adding over $r = 0, 1, \ldots, n$ gives the theorem.

There is a second proof of the theorem by mathematical induction; induction is formative only in this course, so the counting proof above is the one to learn.

The coefficient counts the ways of picking the brackets.
Brackets that give $b$ ( $r$ )	Which brackets, numbered $1$ to $4$	Ways	Term formed
$0$	none	$1$	$a^{4}b^{0}$
$1$	{1} {2} {3} {4}	$4$	$a^{3}b^{1}$
$2$	{1,2} {1,3} {2,3} {1,4} {2,4} {3,4}	$6$	$a^{2}b^{2}$
$3$	{1,2,3} {1,2,4} {1,3,4} {2,3,4}	$4$	$a^{1}b^{3}$
$4$	{1,2,3,4}	$1$	$a^{0}b^{4}$

$(x+y)^{6} = x^{6} + 6x^{5}y + 15x^{4}y^{2} + 20x^{3}y^{3} + 15x^{2}y^{4} + 6xy^{5} + y^{6}$.
$r$	$\binom{6}{r}$	Term $\binom{6}{r}x^{6-r}y^{r}$
$0$	$1$	$x^{6}$
$1$	$6$	$6x^{5}y$
$2$	$15$	$15x^{4}y^{2}$
$3$	$20$	$20x^{3}y^{3}$
$4$	$15$	$15x^{2}y^{4}$
$5$	$6$	$6xy^{5}$
$6$	$1$	$y^{6}$

Formulas

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

Key terms

Binomial theorem: The expansion of $(a+b)^{n}$ as $\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^{r}$ for a positive integer $n$ .

3. Features and special cases

The expansion has $n + 1$ terms, and the powers of $a$ and $b$ add up to $n$ in every term.
Terms equally far from the two ends have equal coefficients.
Putting $a = b = 1$ : $\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n} = 2^{n}$ .
Putting $a = 1$ , $b = -1$ : $\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \cdots = 0$ , so the coefficients in the even places and in the odd places have the same sum, $2^{n-1}$ .
$(a - b)^{n}$ : use $-b$ as the second term, so the signs alternate.
$(a + b)^{n} + (a - b)^{n}$ keeps only the even powers of $b$ and doubles them; $(a + b)^{n} - (a - b)^{n}$ keeps only the odd powers and doubles them.
When a term is a fraction, write $\dfrac{1}{x}$ as $x^{-1}$ and combine the powers.

Put $a = b = 1$: sum $2^{n}$. Put $a = 1$, $b = -1$: sum $0$.
$n$	Coefficients	Sum of the row	Alternating sum
$0$	1	$2^{0} = 1$	$1$
$1$	1, 1	$2^{1} = 2$	$0$
$2$	1, 2, 1	$2^{2} = 4$	$0$
$3$	1, 3, 3, 1	$2^{3} = 8$	$0$
$4$	1, 4, 6, 4, 1	$2^{4} = 16$	$0$
$5$	1, 5, 10, 10, 5, 1	$2^{5} = 32$	$0$
$6$	1, 6, 15, 20, 15, 6, 1	$2^{6} = 64$	$0$
$7$	1, 7, 21, 35, 35, 21, 7, 1	$2^{7} = 128$	$0$
$8$	1, 8, 28, 56, 70, 56, 28, 8, 1	$2^{8} = 256$	$0$

Signs alternate because the second term is negative.
$r$	Term $\binom{4}{r}(3x)^{4-r}(-2y)^{r}$	Simplified
$0$	$\binom{4}{0}(3x)^{4}(-2y)^{0}$	$81x^{4}$
$1$	$\binom{4}{1}(3x)^{3}(-2y)^{1}$	$-216x^{3}y$
$2$	$\binom{4}{2}(3x)^{2}(-2y)^{2}$	$216x^{2}y^{2}$
$3$	$\binom{4}{3}(3x)^{1}(-2y)^{3}$	$-96xy^{3}$
$4$	$\binom{4}{4}(3x)^{0}(-2y)^{4}$	$16y^{4}$
Sum		$81x^{4} - 216x^{3}y + 216x^{2}y^{2} - 96xy^{3} + 16y^{4}$

A negative second term gives alternating signs; $\left(\dfrac{1}{x}\right)^{r} = x^{-r}$.
$r$	Term $\binom{4}{r}x^{4-r}\left(-\dfrac{1}{x}\right)^{r}$	Simplified
$0$	$\binom{4}{0}x^{4}\left(-\dfrac{1}{x}\right)^{0}$	$x^{4}$
$1$	$\binom{4}{1}x^{3}\left(-\dfrac{1}{x}\right)^{1}$	$-4x^{2}$
$2$	$\binom{4}{2}x^{2}\left(-\dfrac{1}{x}\right)^{2}$	$6$
$3$	$\binom{4}{3}x^{1}\left(-\dfrac{1}{x}\right)^{3}$	$-4x^{-2}$
$4$	$\binom{4}{4}x^{0}\left(-\dfrac{1}{x}\right)^{4}$	$x^{-4}$
Sum		$x^{4} - 4x^{2} + 6 - 4x^{-2} + x^{-4}$

Formulas

Sum of the coefficients
$\sum_{r=0}^{n}\binom{n}{r} = 2^{n}$
Alternating sum
$\sum_{r=0}^{n}(-1)^{r}\binom{n}{r} = 0$

Key terms

Alternating signs: Signs that go $+, -, +, -, \ldots$ , which happens when the second term is negative.

4. Numerical applications

Values. Split a number into a round number and a small one, for example $52^{4} = (50 + 2)^{4}$ , and add the terms.

Estimates. For a small $x$ , $(1 + x)^{n} = 1 + nx + \binom{n}{2}x^{2} + \cdots$ ; the terms shrink fast, so a few terms give a good estimate. Check how many decimal places you need before you stop.

Remainders and divisibility. Write the number as a multiple of the divisor plus or minus a small number, $(km \pm 1)^{e}$ . Every term except the last has the factor $m$ , so only the last term matters: $(-1)^{e}$ or $1$ . A remainder must be between $0$ and $m - 1$ , so $-1$ is written as $m - 1$ .

Split the number as a round number plus a small one.
$r$	Term	Value
$0$	$\binom{4}{0} \times 50^{4} \times 2^{0}$	$6250000$
$1$	$\binom{4}{1} \times 50^{3} \times 2^{1}$	$1000000$
$2$	$\binom{4}{2} \times 50^{2} \times 2^{2}$	$60000$
$3$	$\binom{4}{3} \times 50^{1} \times 2^{3}$	$1600$
$4$	$\binom{4}{4} \times 50^{0} \times 2^{4}$	$16$
Total	$52^{4}$	$7311616$

Later terms are tiny: a few terms give a good estimate.
$r$	Term $\binom{6}{r}(0.03)^{r}$	Running sum
$0$	$1$	$1$
$1$	$0.18$	$1.18$
$2$	$0.0135$	$1.1935$
$3$	$0.00054$	$1.19404$
$4$	$0.00001215$	$1.19405215$
$5$	$0.0000001458$	$1.1940522958$
$6$	$0.000000000729$	$1.194052296529$

Write the number as a multiple plus or minus 1.
Number	Written as	The term without the multiple	Remainder
$13^{30} \div 12$	$(12 + 1)^{30}$	$\text{the last term is } 1$	$1$
$8^{20} \div 7$	$(7 + 1)^{20}$	$\text{the last term is } 1$	$1$
$11^{25} \div 12$	$(12 - 1)^{25}$	$\text{the last term is } (-1)^{25} = -1$	$11$
$7^{42} \div 50$	$(50 - 1)^{21}$	$\text{the last term is } (-1)^{21} = -1$	$49$

$7^{42}$ leaves remainder $49$ on division by $50$.
Step	Working
Make a power of $49$	$7^{42} = (7^{2})^{21} = 49^{21}$
Write $49$ next to a multiple of $50$	$49^{21} = (50 - 1)^{21}$
Expand	$\binom{21}{0}50^{21} - \binom{21}{1}50^{20} + \cdots - 1$
Every term but the last has the factor $50$	$7^{42} = 50k - 1$
Remainder	$50k - 1 = 50(k - 1) + 49$ , so the remainder is $49$

Formulas

Three-term estimate for small $x$
$(1 + x)^{n} \approx 1 + nx + \binom{n}{2}x^{2}$

Key terms

Remainder: What is left when a number is divided by another; it lies between $0$ and the divisor minus $1$ .

5. The general term (formative)

In $(a + b)^{n}$ the $(r+1)$ th term is

T_{r+1} = \binom{n}{r}a^{n-r}b^{r}, \quad r = 0, 1, \ldots, n.

For the $k$ th term put $r = k - 1$ . The $(n + 1)$ th term from the beginning is the first term from the end, and the $k$ th term from the end is the $(n - k + 2)$ th term from the beginning. Keep the sign and the coefficient of each term inside the bracket when raising it to the power.

This topic is for formative assessment only; it is taught here but kept light.

The $(r+1)$th term has $b^{r}$ and the coefficient $\binom{n}{r}$.
Term number	$r$	$T_{r+1} = \binom{7}{r}a^{7-r}b^{r}$	Coefficient
$T_{1}$	$0$	$\binom{7}{0}a^{7}b^{0}$	$1$
$T_{2}$	$1$	$\binom{7}{1}a^{6}b^{1}$	$7$
$T_{3}$	$2$	$\binom{7}{2}a^{5}b^{2}$	$21$
$T_{4}$	$3$	$\binom{7}{3}a^{4}b^{3}$	$35$
$T_{5}$	$4$	$\binom{7}{4}a^{3}b^{4}$	$35$
$T_{6}$	$5$	$\binom{7}{5}a^{2}b^{5}$	$21$
$T_{7}$	$6$	$\binom{7}{6}a^{1}b^{6}$	$7$
$T_{8}$	$7$	$\binom{7}{7}a^{0}b^{7}$	$1$

Seven terms for the sixth power.
Term	$r$	$\binom{6}{r}x^{6-r}\,2^{r}$	Simplified
$T_{1}$	$0$	$\binom{6}{0}x^{6}2^{0}$	$x^{6}$
$T_{2}$	$1$	$\binom{6}{1}x^{5}2^{1}$	$12x^{5}$
$T_{3}$	$2$	$\binom{6}{2}x^{4}2^{2}$	$60x^{4}$
$T_{4}$	$3$	$\binom{6}{3}x^{3}2^{3}$	$160x^{3}$
$T_{5}$	$4$	$\binom{6}{4}x^{2}2^{4}$	$240x^{2}$
$T_{6}$	$5$	$\binom{6}{5}x^{1}2^{5}$	$192x$
$T_{7}$	$6$	$\binom{6}{6}x^{0}2^{6}$	$64$

The fourth term has $r = 3$.
Step	Working
General term	$T_{r+1} = \binom{8}{r}(3x)^{8-r}\left(-\dfrac{1}{2x}\right)^{r}$
Fourth term	$r + 1 = 4$ , so $r = 3$
Substitute	$\binom{8}{3}(3x)^{5}\left(-\dfrac{1}{2x}\right)^{3}$
Simplify	$56 \times 243x^{5} \times \left(-\dfrac{1}{8x^{3}}\right)$
Value	$T_{4} = -1701x^{2}$

Formulas

General term
$T_{r+1} = \binom{n}{r}a^{n-r}b^{r}$

Key terms

General term: $T_{r+1} = \binom{n}{r}a^{n-r}b^{r}$ , the term with $b^{r}$ .

6. A particular term and the term independent of x (formative)

To find the term containing $x^{k}$ :

Write $T_{r+1}$ and collect the power of $x$ as an expression in $r$ .
Set it equal to $k$ and solve for $r$ .
If $r$ is a whole number with $0 \le r \le n$ , substitute it; otherwise there is no such term.

For the term independent of $x$ take $k = 0$ . Example: in $\left(x^{3} + \dfrac{2}{x^{2}}\right)^{10}$ the power of $x$ is $30 - 5r$ , zero at $r = 6$ , and the term is $\binom{10}{6}2^{6} = 13440$ .

Always check that $r$ is a whole number between $0$ and $n$.

The power is $30 - 5r$; it is $0$ at $r = 6$.
$r$	Power of $x$ in $T_{r+1}$	Coefficient $\binom{10}{r}2^{r}$
$0$	$30$	$1$
$1$	$25$	$20$
$2$	$20$	$180$
$3$	$15$	$960$
$4$	$10$	$3360$
$5$	$5$	$8064$
$6$	$0$	$13440$ (independent of $x$ )
$7$	$-5$	$15360$
$8$	$-10$	$11520$
$9$	$-15$	$5120$
$10$	$-20$	$1024$

Solve for $r$, then substitute.
Step	Working
General term	$T_{r+1} = \binom{10}{r}(x^{3})^{10-r}\left(\dfrac{2}{x^{2}}\right)^{r} = \binom{10}{r}2^{r}x^{30-5r}$
Independent of $x$	$30 - 5r = 0$ , so $r = 6$
Value	$\binom{10}{6}2^{6} = 210 \times 64 = 13440$

Formulas

Condition for $x^{k}$
$\text{power of } x \text{ in } T_{r+1} = k$

Key terms

Term independent of $x$: The term in which the power of $x$ is $0$ (the constant term).

7. Middle terms and equal coefficients (formative)

$n$ even: one middle term, $T_{\frac{n}{2}+1}$ , with the greatest coefficient $\binom{n}{n/2}$ .
$n$ odd: two middle terms, $T_{\frac{n+1}{2}}$ and $T_{\frac{n+3}{2}}$ , whose binomial coefficients are equal.
$\binom{n}{r} = \binom{n}{s}$ with $r \ne s$ gives $r + s = n$ .

The binomial coefficients rise to the middle and fall again, so the middle term has the greatest binomial coefficient (not always the greatest term, because the numbers inside the bracket also matter).

the binomial coefficients of the tenth power joined by a line

The middle term of $(a+b)^{10}$ has $r = 5$: $\binom{10}{5} = 252$.

Odd power: the two middle terms are $T_{4}$ and $T_{5}$.
Term	Working	Value
$T_{4}$	$\binom{7}{3}x^{4}(2y)^{3} = 35 \times 8\,x^{4}y^{3}$	$280x^{4}y^{3}$
$T_{5}$	$\binom{7}{4}x^{3}(2y)^{4} = 35 \times 16\,x^{3}y^{4}$	$560x^{3}y^{4}$

$\binom{n}{r} = \binom{n}{s}$ with $r \ne s$ means $r + s = n$.
Step	Working
The coefficients of $x^{7}$ and $x^{13}$	$\binom{n}{7}$ and $\binom{n}{13}$
Equal binomial coefficients with different lower numbers	$7 + 13 = n$
Value	$n = 20$
Check	$\binom{20}{7} = \binom{20}{13} = 77520$

Formulas

Equal coefficients
$\binom{n}{r} = \binom{n}{s},\ r \ne s \ \Rightarrow\ r + s = n$

Key terms

Middle term: The term at the centre of the expansion: one term for an even power, two for an odd power.

Exam tips

How marks are awarded

Expansion of a sum or difference (2–5 marks): the row or $\binom{n}{r}$ for the coefficients, the substitution of both terms (sign and number inside the bracket), the simplified terms.
Numerical value or estimate (2–3 marks): the split of the number, the terms listed, the total; the stated number of decimal places.
Remainder or divisibility (3–5 marks): the nearby multiple, the term without the factor, the remainder written between $0$ and the divisor.
Proof of the theorem or of a sum of coefficients (3–5 marks): the choice of one letter from each bracket, $\binom{n}{r}$ ways, the sum over $r$ ; or the substitutions $a = b = 1$ and $a = 1, b = -1$ .
General, particular and middle terms (2–5 marks, formative): the general term, the value of $r$ , the simplified term; even or odd first for middle terms.

Common mistakes

Losing the sign of a negative second term.
Not raising the number in front of a letter to the power, as in $(3x)^{4}$ .
Writing $(a+b)^{n} = a^{n} + b^{n}$ .
Using the term number as $r$ (the $k$ th term has $r = k - 1$ ).
Accepting a fractional $r$ for a term containing a given power of $x$ .
Forgetting that $n$ even has one middle term and $n$ odd has two.
Leaving a remainder as $-1$ instead of the divisor minus $1$ .

1. Pascal's triangle

Formulas

Key terms

2. The binomial theorem and its proof

Formulas

Key terms

3. Features and special cases

Formulas

Key terms

4. Numerical applications

Formulas

Key terms

5. The general term (formative)

Formulas

Key terms

6. A particular term and the term independent of x (formative)

Formulas

Key terms

7. Middle terms and equal coefficients (formative)

Formulas

Key terms

Exam tips

Cookies