Binomial Theorem Notes

Binomial Theorem

8.1 Overview

8.1.1 Binomial Expression

An expression consisting of two terms connected by a $+$ or $-$ sign is called a binomial expression. Examples include:

• $x + a$

• $2x - 3y$

• $x + \frac{1}{x}$

• $\frac{x}{y} - \frac{y}{x}$

8.1.2 Binomial Theorem

If $a$ and $b$ are real numbers and $n$ is a positive integer, then
$(a + b)^n = {}^nC<em>0 a^n + {}^nC</em>1 a^{n-1} b^1 + {}^nC<em>2 a^{n-2} b^2 + … + {}^nC</em>r a^{n-r} b^r + … + {}^nC<em>n b^n$ where ${}^nC</em>r = \frac{n!}{r!(n-r)!}$ for $0 \le r \le n$.

• General Term: The general term or $(r + 1)^{th}$ term in the expansion is given by
  $T<em>{r+1} = {}^nC</em>r a^{n-r} b^r$

8.1.3 Important Observations

1. Total Number of Terms: The total number of terms in the binomial expansion of $(a + b)^n$ is $n + 1$, i.e., one more than the exponent $n$.

2. Powers of a and b: In the expansion, the power of $a$ decreases by one in each subsequent term, while the power of $b$ increases by one, until the power of $b$ equals the power of the binomial. Specifically:

   • The power of $a$ starts at $n$ in the first term and ends at $0$ in the last term.

   • The power of $b$ starts at $0$ in the first term and ends at $n$ in the last term.

3. Sum of Indices: In any term, the sum of the exponents of $a$ and $b$ is equal to $n$, the power of the binomial.

4. Pascal's Triangle: The coefficients in the expansion follow a pattern known as Pascal's triangle.

Pascal's Triangle

Index of Binomial Coefficient of various terms

0                                   1
1                             1           1
2                         1       2          1
3                     1      3       3          1
4                 1     4       6       4       1
5            1    5       10      10      5    1
Each coefficient of any row is obtained by adding two coefficients in the preceding row, one on the immediate left and the other on the immediate right and each row is bounded by 1 on both sides.
The (r + 1)th term or general term is given by Tr + 1 = nCr an – r br

8.1.4 Particular Cases

If $n$ is a positive integer, then

$(a + b)^n = {}^nC<em>0 a^n b^0 + {}^nC</em>1 a^{n-1} b^1 + {}^nC<em>2 a^{n-2} b^2 + … + {}^nC</em>r a^{n-r} b^r + … + {}^nC_n a^0 b^n$

1. Replacing b by -b:

$(a - b)^n = {}^nC<em>0 a^n b^0 - {}^nC</em>1 a^{n-1} b^1 + {}^nC<em>2 a^{n-2} b^2 + … + (-1)^r {}^nC</em>r a^{n-r} b^r + … + (-1)^n {}^nC_n a^0 b^n$

2. Adding (1) and (2):

$(a + b)^n + (a - b)^n = 2 [ {}^nC<em>0 a^n b^0 + {}^nC</em>2 a^{n-2} b^2 + {}^nC_4 a^{n-4} b^4 + … ] = 2 [ \text{terms at odd places} ]$

3. Subtracting (2) from (1):

$(a + b)^n - (a - b)^n = 2 [ {}^nC<em>1 a^{n-1} b^1 + {}^nC</em>3 a^{n-3} b^3 + … ] = 2 [ \text{sum of terms at even places} ]$

4. Replacing a by 1 and b by x:

$(1 + x)^n = {}^nC<em>0 x^0 + {}^nC</em>1 x + {}^nC<em>2 x^2 + … + {}^nC</em>r x^r + … + {}^nC<em>{n-1} x^{n-1} + {}^nC</em>n x^n$

$(1 + x)^n = \sum<em>{r=0}^n {}^nC</em>r x^r$

5. Replacing a by 1 and b by -x:

$(1 - x)^n = {}^nC<em>0 x^0 - {}^nC</em>1 x + {}^nC<em>2 x^2 + … + {}^nC</em>{n-1} (-1)^{n-1} x^{n-1} + {}^nC_n (-1)^n x^n$

$(1 - x)^n = \sum<em>{r=0}^n (-1)^r {}^nC</em>r x^r$

8.1.5 The pth Term from the End

The $p^{th}$ term from the end in the expansion of $(a + b)^n$ is $(n - p + 2)^{th}$ term from the beginning.

8.1.6 Middle Terms

The middle term depends upon the value of n.

• (a) If n is even:
  The total number of terms in the expansion of $(a + b)^n$ is $n + 1$ (odd). Hence, there is only one middle term, i.e., the $\left( \frac{n}{2} + 1 \right)^{th}$ term is the middle term.

• (b) If n is odd:
  The total number of terms in the expansion of $(a + b)^n$ is $n + 1$ (even). So there are two middle terms i.e., the $\left( \frac{n+1}{2} \right)^{th}$ and $\left( \frac{n+3}{2} \right)^{th}$ terms are two middle terms.

8.1.7 Binomial Coefficient

In the Binomial expression, we have

$(a + b)^n = {}^nC<em>0 a^n + {}^nC</em>1 a^{n-1} b + {}^nC<em>2 a^{n-2} b^2 + … + {}^nC</em>n b^n$
The coefficients ${}^nC<em>0, {}^nC</em>1, {}^nC<em>2, …, {}^nC</em>n$ are known as binomial or combinatorial coefficients.
Putting $a = b = 1$ in (1), we get

${}^nC<em>0 + {}^nC</em>1 + {}^nC<em>2 + … + {}^nC</em>n = 2^n$
Thus the sum of all the binomial coefficients is equal to $2^n$.
Again, putting $a = 1$ and $b = -1$ in (i), we get

${}^nC<em>0 + {}^nC</em>2 + {}^nC<em>4 + … = {}^nC</em>1 + {}^nC<em>3 + {}^nC</em>5 + …$

${}^nC<em>0 + {}^nC</em>2 + {}^nC<em>4 + … = {}^nC</em>1 + {}^nC<em>3 + {}^nC</em>5 + … = 2^{n-1}$

8.2 Solved Examples

Example 1

Find the $r^{th}$ term in the expansion of $\left( x + \frac{1}{x} \right)^{2r}$.

Solution:

We have

$T<em>r = {}^{2r}C</em>{r-1} (x)^{2r - (r-1)} \left( \frac{1}{x} \right)^{r-1} = {}^{2r}C<em>{r-1} x^{r+1} x^{-(r-1)} = {}^{2r}C</em>{r-1} x^2$

Example 2

Expand the following $(1 - x + x^2)^4$

Solution:

Put $1 - x = y$. Then

$(1 - x + x^2)^4 = (y + x^2)^4 = {}^4C<em>0 y^4 (x^2)^0 + {}^4C</em>1 y^3 (x^2)^1 + {}^4C<em>2 y^2 (x^2)^2 + {}^4C</em>3 y (x^2)^3 + {}^4C_4 (x^2)^4$

$= y^4 + 4y^3 x^2 + 6y^2 x^4 + 4y x^6 + x^8$

$= (1 - x)^4 + 4x^2 (1 - x)^3 + 6x^4 (1 - x)^2 + 4x^6 (1 - x) + x^8$

$= 1 - 4x + 10x^2 - 16x^3 + 19x^4 - 16x^5 + 10x^6 - 4x^7 + x^8$

Example 3

Find the 4th term from the end in the expansion of $\left( \frac{x^2}{2} - \frac{2}{x} \right)^9$

Solution:

Since $r^{th}$ term from the end in the expansion of $(a + b)^n$ is $(n - r + 2)^{th}$ term from the beginning.

Therefore 4th term from the end is $9 - 4 + 2$, i.e., 7th term from the beginning, which is given by

$T<em>7 = {}^9C</em>6 \left( \frac{x^2}{2} \right)^{9-6} \left( -\frac{2}{x} \right)^6 = {}^9C_6 \left( \frac{x^6}{8} \right) \left( \frac{64}{x^6} \right) = \frac{9!}{6!3!} \cdot \frac{64}{8} \cdot \frac{x^6}{x^6} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \cdot 8 = 672$

Example 4

Evaluate: $(x^2 - \sqrt{1-x^2})^4 + (x^2 + \sqrt{1-x^2})^4$

Solution:

Putting $y = \sqrt{1-x^2}$, we get

The given expression $= (x^2 - y)^4 + (x^2 + y)^4 = 2 [ x^8 + {}^4C<em>2 x^4 y^2 + {}^4C</em>4 y^4 ]$

$= 2 [ x^8 + 6 x^4 (1 - x^2) + (1 - x^2)^2 ]$

$= 2 [ x^8 + 6x^4 (1-x^2) + (1 - 2x^2 + x^4) ] = 2x^8 - 12x^6 + 14x^4 - 4x^2 + 2$

Example 5

Find the coefficient of $x^{11}$ in the expansion of $\left( x^3 - \frac{2}{x^2} \right)^{12}$

Solution:

Let the general term, i.e., $(r + 1)^{th}$ contain $x^{11}$. We have

$T<em>{r+1} = {}^{12}C</em>r (x^3)^{12-r} \left( -\frac{2}{x^2} \right)^r = {}^{12}C_r x^{36-3r} (-1)^r (2^r) x^{-2r}$

$= {}^{12}C_r (-1)^r 2^r x^{36-5r}$

Now for this to contain $x^{11}$, we observe that $36 - 5r = 11$, i.e., $r = 5$

Thus, the coefficient of $x^{11}$ is ${}^{12}C_5 (-1)^5 2^5 = -25344$

Example 6

Determine whether the expansion of $\left( x^2 - \frac{2}{x} \right)^{18}$ will contain a term containing $x^{10}$?

Solution:

Let $T_{r+1}$ contain $x^{10}$. Then

$T<em>{r+1} = {}^{18}C</em>r (x^2)^{18-r} \left( -\frac{2}{x} \right)^r = {}^{18}C_r x^{36-2r} (-1)^r 2^r x^{-r}$

$= (-1)^r 2^r {}^{18}C_r x^{36-3r}$

Thus, $36 - 3r = 10$, i.e., $r = \frac{26}{3}$

Since r is a fraction, the given expansion cannot have a term containing $x^{10}$.

Example 7

Find the term independent of $x$ in the expansion of $\left( \frac{2}{3} x^2 + \frac{3}{2x} \right)^{10}$.

Solution:

Let $(r + 1)^{th}$ term be independent of $x$ which is given by

$T<em>{r+1} = {}^{10}C</em>r \left( \frac{2}{3} x^2 \right)^{10-r} \left( \frac{3}{2x} \right)^r = {}^{10}C_r \left( \frac{2}{3} \right)^{10-r} x^{2(10-r)} \left( \frac{3}{2} \right)^r x^{-r}$

$= {}^{10}C<em>r \left( \frac{2}{3} \right)^{10-r} \left( \frac{3}{2} \right)^r x^{20 - 2r - r} = {}^{10}C</em>r \left( \frac{2}{3} \right)^{10-r} \left( \frac{3}{2} \right)^r x^{20 - 3r}$

Since the term is independent of $x$, we have

$20 - 3r = 0 \implies r = \frac{20}{3}$

Therefore, there is likely a typo in the example and a solution can not be found.

Example 8

Find the middle term in the expansion of $\left( \frac{a}{x} - bx^2 \right)^{12}$.

Solution:

Since the power of binomial is even, it has one middle term which is the $\left( \frac{12}{2} + 1 \right)^{th}$ term and it is given by

$T<em>7 = {}^{12}C</em>6 \left( \frac{a}{x} \right)^{12-6} (-bx^2)^6 = {}^{12}C<em>6 a^6 x^{-6} b^6 x^{12} = {}^{12}C</em>6 a^6 b^6 x^6$

Example 9

Find the middle term (terms) in the expansion of $\left( x - \frac{p}{x} \right)^{9}$.

Solution:

Since the power of binomial is odd. Therefore, we have two middle terms which are 5th and 6th terms. These are given by

$T<em>5 = {}^9C</em>4 (x)^{9-4} \left( -\frac{p}{x} \right)^4 = 126 p^4 x$

$T<em>6 = {}^9C</em>5 (x)^{9-5} \left( -\frac{p}{x} \right)^5 = -126 \frac{p^5}{x}$

Example 10

Show that $2^{4n+4} - 15n - 16$, where $n \in N$ is divisible by 225.

Solution:

We have

$2^{4n+4} - 15n - 16 = 2^4 (n+1) - 15n - 16 = 16^{n+1} - 15n - 16 = (1 + 15)^{n+1} - 15n - 16$

$= {}^{n+1}C<em>0 15^0 + {}^{n+1}C</em>1 15^1 + {}^{n+1}C<em>2 15^2 + {}^{n+1}C</em>3 15^3 + … + {}^{n+1}C_{n+1} (15)^{n+1} - 15n - 16$

$= 1 + (n+1)15 + {}^{n+1}C<em>2 15^2 + {}^{n+1}C</em>3 15^3 + … + {}^{n+1}C_{n+1} (15)^{n+1} - 15n - 16$

$= 1 + 15n + 15 + {}^{n+1}C<em>2 15^2 + {}^{n+1}C</em>3 15^3 + … + {}^{n+1}C_{n+1} (15)^{n+1} - 15n - 16$

$= 15^2 [ {}^{n+1}C<em>2 + {}^{n+1}C</em>3 15 + … \text{ so on} ]$

Thus, $2^{4n+4} - 15n - 16$ is divisible by 225.

Example 11

Find numerically the greatest term in the expansion of $(2 + 3x)^9$, where $x = \frac{3}{2}$.

Solution:

$(2 + 3x)^9 = 2^9 \left( 1 + \frac{3}{2} x \right)^9$

Now,

$\frac{T<em>{r+1}}{T</em>r} = \frac{{}^9C<em>r 2^{9-r} (3x)^r}{{}^9C</em>{r-1} 2^{9-r+1} (3x)^{r-1}} = \frac{{}^9C<em>r}{{}^9C</em>{r-1}} \frac{3x}{2} = \frac{9-r+1}{r} \frac{3x}{2} = \frac{10-r}{r} \frac{3x}{2}$

Since $x = \frac{3}{2}$,

$\frac{T<em>{r+1}}{T</em>r} = \frac{10-r}{r} \cdot \frac{9}{4}$

$\frac{T<em>{r+1}}{T</em>r} \ge 1 \implies \frac{10-r}{r} \cdot \frac{9}{4} \ge 1 \implies 90 - 9r \ge 4r \implies r \le \frac{90}{13} \implies r \le 6 \frac{12}{13}$

Thus the maximum value of $r$ is 6. Therefore, the greatest term is $T<em>{r+1} = T</em>7$.

$T<em>7 = {}^9C</em>6 2^{9-6} (3x)^6 = {}^9C_6 2^3 \left( \frac{9}{4} \right)^3 = 84 \cdot 8 \cdot \frac{729}{64} = \frac{15309}{2}$

Example 12

If $n$ is a positive integer, find the coefficient of $x^{-1}$ in the expansion of $(1 + x)^n \left( 1 + \frac{1}{x} \right)^n$.

Solution:

We have

$\left( 1 + x \right)^n \left( 1 + \frac{1}{x} \right)^n = \left( \frac{x+1}{x} \right)^n \left( 1 + x \right)^n = \frac{\left( 1 + x \right)^{2n}}{x^n}$

Now to find the coefficient of $x^{-1}$ in $(1 + x)^n \left( 1 + \frac{1}{x} \right)^n$, it is equivalent to finding coefficient of $x^{-1}$ in $\frac{\left( 1 + x \right)^{2n}}{x^n}$ which in turn is equal to the coefficient of $x^{n-1}$ in the expansion of $(1 + x)^{2n}$.
$(1 + x)^{2n} = {}^{2n}C<em>0 x^0 + {}^{2n}C</em>1 x^1 + {}^{2n}C<em>2 x^2 + … + {}^{2n}C</em>{n-1} x^{n-1} + … + {}^{2n}C<em>{2n} x^{2n}$ Thus the coefficient of $x^{n-1}$ is ${}^{2n}C</em>{n-1} = \frac{(2n)!}{(n-1)! (2n-n+1)!} = \frac{(2n)!}{(n-1)!(n+1)!} = \frac{2n!}{(n+1)(n!)(n-1)!} = \frac{2n}{n+1} \cdot \frac{n!}{(n-1)!}$

Example 13

Which of the following is larger? $99^{50} + 101^{50}$ or $100^{50}$

We have

$(101)^{50} = (100 + 1)^{50} = 100^{50} + 50 (100)^{49} + \frac{50 \cdot 49}{2 \cdot 1} (100)^{48} + \frac{50 \cdot 49 \cdot 48}{3 \cdot 2 \cdot 1} (100)^{47} + …$

$(99)^{50} = (100 - 1)^{50} = 100^{50} - 50 (100)^{49} + \frac{50 \cdot 49}{2 \cdot 1} (100)^{48} - \frac{50 \cdot 49 \cdot 48}{3 \cdot 2 \cdot 1} (100)^{47} + …$

(101)^{50} + (99)^{50} = 2 [ 100^{50} + \frac{50 \cdot 49}{2 \cdot 1} (100)^{48} + … ] > 100^{50}

Hence 101^{50} > 99^{50} + 100^{50}

Example 14

Find the coefficient of $x^{50}$ after simplifying and collecting the like terms in the expansion of $(1 + x)^{1000} + x (1 + x)^{999} + x^2 (1 + x)^{998} + … + x^{1000}$.

Solution:

Since the above series is a geometric series with the common ratio $\frac{x}{1+x}$, its sum is

$\frac{(1+x)^{1001} - x^{1001}}{(1+x) - x} = (1+x)^{1001} - x^{1001}$

Thus the coefficient of $x^{50}$ is given by

${}^{1001}C_{50}$

Example 15

If $a<em>1, a</em>2, a<em>3$ and $a</em>4$ are the coefficient of any four consecutive terms in the expansion of $(1 + x)^n$, prove that

$\frac{a<em>1}{a</em>1 + a<em>2} + \frac{a</em>3}{a<em>3 + a</em>4} = \frac{2 a<em>2}{a</em>2 + a_3}$

Solution:

Let $a<em>1, a</em>2, a<em>3$ and $a</em>4$ be the coefficient of four consecutive terms $T<em>{r+1}, T</em>{r+2}, T<em>{r+3},$ and $T</em>{r+4}$ respectively.
Then

$a<em>1$ = coefficient of $T</em>{r+1} = {}^nC<em>r$ $a</em>2$ = coefficient of $T<em>{r+2} = {}^nC</em>{r+1}$
$a<em>3$ = coefficient of $T</em>{r+3} = {}^nC<em>{r+2}$ $a</em>4$ = coefficient of $T<em>{r+4} = {}^nC</em>{r+3}$

Thus

$\frac{a<em>1}{a</em>1 + a<em>2} = \frac{{}^nC</em>r }{{}^nC<em>r + {}^nC</em>{r+1}} = \frac{{}^nC<em>r }{{}^{n+1}C</em>{r+1}} = \frac{n! r! (n-r)!}{n+1! (r+1)! (n-r)!} = \frac{r+1}{n+1}$

Similarly,

$\frac{a<em>3}{a</em>3 + a<em>4} = \frac{{}^nC</em>{r+2} }{{}^nC<em>{r+2} + {}^nC</em>{r+3}} = \frac{{}^nC<em>{r+2} }{{}^{n+1}C</em>{r+3}} = \frac{n-r-1}{n+1}$

L.H.S. = $\frac{r+1}{n+1} + \frac{n-r-1}{n+1} = \frac{n}{n+1}$
R.H.S. = $\frac{2a<em>2}{a</em>2+a<em>3} = \frac{2 {}^{n}C</em>{r+1}}{{}^{n}C<em>{r+1}+{}^{n}C</em>{r+2}} = \frac{n}{n+1}$

L.H.S. = R.H.S

Objective Type Questions

Example 16

The total number of terms in the expansion of $(x + a)^{51} - (x - a)^{51}$ after simplification is

(a) 102 (b) 25 (c) 26 (d) None of these

Solution: C is the correct choice since the total number of terms are 52 of which 26 terms get cancelled.

Example 17

If the coefficients of $x^7$ and $x^8$ in $\left( 2 + \frac{x}{3} \right)^n$ are equal, then n is
(a) 56 (b) 55 (c) 45 (d) 15

Solution: B is the correct choice. Since $T<em>{r+1} = {}^nC</em>r a^{n-r} x^r$ in expansion of $(a + x)^n$,
Therefore,

$T<em>8 = {}^nC</em>7 (2)^{n-7} \left( \frac{x}{3} \right)^7 = {}^nC_7 2^{n-7} \frac{x^7}{3^7}$

and

$T<em>9 = {}^nC</em>8 (2)^{n-8} \left( \frac{x}{3} \right)^8 = {}^nC_8 2^{n-8} \frac{x^8}{3^8}$

${}^nC<em>7 \frac{2^{n-7}}{3^7} = {}^nC</em>8 \frac{2^{n-8}}{3^8}$

==> n = 55

Example 18

If $(1 - x + x^2)^n = a<em>0 + a</em>1 x + a<em>2 x^2 + … + a</em>{2n} x^{2n}$, then $a<em>0 + a</em>2 + a<em>4 + … + a</em>{2n}$ equals.
(A) $\frac{3^n + 1}{2}$ (B) $\frac{3^n - 1}{2}$ (C) $\frac{1 - 3^n}{2}$ (D) $\frac{1 + 3^n}{2}$

Solution: A is the correct choice.
Putting $x = 1$ and $-1$ in

$(1 - x + x^2)^n = a<em>0 + a</em>1 x + a<em>2 x^2 + … + a</em>{2n} x^{2n}$

we get

$1 = a<em>0 + a</em>1 + a<em>2 + a</em>3 + … + a_{2n}$

$3^n = a<em>0 - a</em>1 + a<em>2 - a</em>3 + … + a_{2n}$

Adding (1) and (2), we get
$3^n + 1 = 2 (a<em>0 + a</em>2 + a<em>4 + … + a</em>{2n})$

Therefore
$a<em>0 + a</em>2 + a<em>4 + … + a</em>{2n} = \frac{3^n + 1}{2}$

Exercises

The solutions to the exercises can be created upon user request.