Binomial Theorem - Formula, Expansion, Problems

The Binomial Theorem is an important concept of algebra that helps to expand the expressions. A Binomial is an expression with two terms. It is difficult to solve the powers manually therefore this expression makes it simpler to solve. This theorem is widely used in real-life applications in mathematics including calculus etc.

Binomial Expression

An algebraic expression consisting of only two terms is called a Binomial Expression

$
e g \cdot(a+b)^2,\left(\sqrt{x}+\frac{k}{x^2}\right)^5,(x+9 y)^{-2 / 3}
$

If we wanted to expand $(x+y)^{52}$, we might multiply $(x+y)$ by itself fiftytwo times. This could take hours!

But if we examine some simple expansions, we can find patterns that will lead us to a shortcut for finding more complicated binomial expansions.

$\begin{aligned}
& (x+y)^2=x^2+2 x y+y^2 \\
& (x+y)^3=x^3+3 x^2 y+3 x y^2+y^3 \\
& (x+y)^4=x^4+4 x^3 y+6 x^2 y^2+4 x y^3+y^4
\end{aligned}
$

On examining the exponents, we find that with each successive term, the exponent for $x$ decreases by 1 and the exponent for $y$ increases by 1 . The sum of the two exponents is n for each term.

Also the coefficients for $(x+y)^n$ are equal to

$
\begin{aligned}
& \binom{n}{0},\binom{n}{1},\binom{n}{2}, \ldots,\binom{n}{n} \\
& \text { where, }\binom{n}{r}=C(n, r)={ }^n C_r=\frac{n!}{r!(n-r)!}
\end{aligned}
$
These patterns lead us to the Binomial Theorem, which can be used to expand any binomial expression.

For any natural number $n$, binomial expansion is

$
\begin{aligned}
& (x+y)^n={ }^n C_0 x^n+{ }^n C_1 x^{n-1} y+{ }^n C_2 x^{n-2} y^2+\cdots+{ }^n C_n y^n \text { where, } n \in \mathbb{N} \\
= & \sum_{r=0}^n\left({ }^n C_r\right) x^{n-r} y^r
\end{aligned}
$

The combination $\binom{n}{r}$ or ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}$ is called a binomial coefficient.

Note: n can be any rational number for binomial expansion, but formula for n that are not natural numbers is different and will be covered in later sections.

Properties of Binomial Theorem and Binomial Coefficient

1. If n is a natural number, then the expansion $(x+y)^{\mathrm{n}}$ has $\mathrm{n}+1$ terms.
2. In each term, the sum of the index of ' $x$ ' and ' $y$ ' is equal to ' $n$ '.
3. The sum of two consecutive binomial coefficients,

$
{ }^n C_r={ }^n C_{n-r}=\frac{n!}{r!(n-r)!}
$

4. So, if ${ }^n C_x={ }^n C_y$, then either $x=y$ or $n=x+y$
5. Binomial coefficients of the term equidistant from the beginning and end are equal Eg, second binomial coefficient from start $={ }^{\mathrm{n}} \mathrm{C}_1$, and second binomial coeffcient from the end $={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-1}$ These are equal as $1+(n-1)=n$
6.

$
\begin{aligned}
& { }^n C_r=\frac{n}{r} \cdot{ }^{n-1} C_{r-1} \\
& { }^n C_r=\frac{n!}{r!(n-r)!}=\frac{n \cdot(n-1)!}{r \cdot(r-1)!(n-r)!}=\frac{n}{r} \cdot{ }^{n-1} C_{r-1} \\
& \text { 7. } \frac{{ }^n C_r}{{ }^n C_{r-1}}=\frac{n-r+1}{r}
\end{aligned}
$

Some Standard Binomial Expansions:

We know the binomial expansion,

$\begin{aligned}
& (x+y)^{n} = { }^{n} C_0 x^{n} + { }^{n} C_1 x^{n-1} y + { }^{n} C_2 x^{n-2} y^{2} + \cdots + { }^{n} C_{n} y^{n} \\
& \text { i.e. } \quad (x+y)^{n} = \sum_{r=0}^{n} { }^{n} C_{r} x^{n-r} y^{r}
\end{aligned}$

1. Replace ' $y$ ' with ' $-y$ ' in the binomial expansion, we get

$(x-y)^{n} = { }^{n} C_0 x^{n} - { }^{n} C_1 x^{n-1} y + { }^{n} C_2 x^{n-2} y^{2} + \cdots + (-1)^{r} { }^{n} C_{r} x^{n-r} y^{r} + \cdots + (-1)^{n} {\ }^{n} C_{n} y^{n}$

or $(x-y)^{n} = \displaystyle \sum_{r=0}^{n} (-1)^{r} {\ }^{n} C_{r} x^{n-r} y^{r}$

2. In the binomial expansion, $(x+y)^{n}$ replace ' $x$ ' by $1$ and ' $y$ ' by $x$

$(1+x)^{n} = { }^{n} C_0 x^{0} + { }^{n} C_1 x^{1} + { }^{n} C_2 x^{2} + \cdots + { }^{n} C_{r} x^{r} + \cdots + { }^{n} C_{n} x^{n}$
or $(1+x)^{n} = \displaystyle \sum_{r=0}^{n} { }^{n} C_{r} x^{r}$

3. In the binomial expansion, $(x+y)^{n}$ replace ' $x$ ' by ' $1$ ' and ' $y$ ' by ' $-x$ '

$(1-x)^{n} = { }^{n} C_0 x^{0} - { }^{n} C_1 x^{1} + { }^{n} C_2 x^{2} - \cdots + (-1)^{r} { }^{n} C_{r} x^{r} + \cdots + (-1)^{n} { \ }^{n} C_{n} x^{n}$

or $(1-x)^{n} = \displaystyle \sum_{r=0}^{n} (-1)^{r} {\ }^{n} C_{r} x^{r}$

4. Addition: $(x+y)^{n} + (x-y)^{n}$

$(x+y)^{n} + (x-y)^{n} = 2 \left[ { }^{n} C_0 x^{n} y^{0} + { }^{n} C_2 x^{n-2} y^{2} + { }^{n} C_4 x^{n-4} y^{4} + \ldots \right]$

If ' $n$ ' is odd, then the number of terms is $\frac{n+1}{2}$.

If ' $n$ ' is even, then the number of terms is $\frac{n}{2} + 1$.

5. Subtraction: $(x+y)^{n} - (x-y)^{n}$

$(x+y)^{n} - (x-y)^{n} = 2 \left[ { }^{n} C_1 x^{n-1} y^{1} + { }^{n} C_3 x^{n-3} y^{3} + { }^{n} C_5 x^{n-5} y^{5} + \ldots \right]$

If ' $n$ ' is odd, then the number of terms is $\frac{n+1}{2}$.

If ' $n$ ' is even, then the number of terms is $\frac{n}{2}$.

Solved Examples Based on Binomial Theorem:

Example 1: Which of the following is NOT a binomial expression?

1) $(x+y)^{-5}$
2) $(x-y)^7$
3) $(a+b)^{\frac{-8}{5}}$
4) $(3 a+b)^i$

Solution:

We can not have complex numbers in exponents.

Only positive and negative integers and rational numbers are allowed.

Hence, the answer is the option 4.

Example 2: For natural numbers $m, n \space{\text {if }}(1-y)^m(1+y)^n=1+a_1 y+a_2 y^2+\ldots \ldots .$, and $a_1=a_2=10$, then $(m, n)$ is :

1) (20, 45)

2) (35, 20)

3) (45, 35)

4) (35, 45).

Solution:

As we learned in

Expression of Binomial Theorem -

$
(x+a)^n={ }^n C_0 x^n a^0+{ }^n C_1 x^{n-1} a^1+{ }^n C_2 x^{n-2} a^2+----{ }^n C_n x^0 a^n
$
where
for n is +ve integer
Now,
$
\begin{aligned}
& (1-y)^m(1+y)^n \\
& \Rightarrow\left(1-{ }^m C_1 y+{ }^m C_2 y^2 \ldots \ldots . .\right)\left(1+{ }^n C_1 y+{ }^n C_2 y^2 \ldots \ldots . .\right) \\
& \Rightarrow\left(1+(n-m) y+\left({ }^m C_2+{ }^n C_2-m n\right) y^2 \ldots \ldots \ldots .\right)
\end{aligned}
$

So $n-m=10$
$
\text { and } \frac{m(m-1)}{2}+\frac{n(n-1)}{2}-m n=10
$

Using n = m + 10 from the first equation and substituting it in the second, we get values of (m, n) satisfying (35, 45)

Hence, the answer is an option (4).

Example 3: The expansion of $\left(x+\frac{1}{x}\right)^n$ is:

1) $\displaystyle \sum_{r=0}^n{ }^n C_r x^{n-2 r}$

2) $\displaystyle \sum_{r=2}^n{ }^n C_r x^{n-2 r}$

3) $\displaystyle \sum_{r=0}^n\left({ }^n C_r x^{n-r}\right)$

4) None of these

Solution:

Binomial Theorem
$
\begin{aligned}
(x+y)^n & =\sum_{r=0}^n\binom{n}{r} x^{n-r} y^r \\
& =x^n+\binom{n}{1} x^{n-1} y+\binom{n}{2} x^{n-2} y^2+\ldots+\binom{n}{n-1} x y^{n-1}+y^n
\end{aligned}
$

Now,

$\begin{aligned}\left(x+\frac{1}{x}\right)^n & =\sum_{r=0}^n{ }^n C_r x^{n-r} \cdot\left(\frac{1}{x}\right)^r \\ & =\sum_{r=0}^n{ }^n C_r x^{n-r} x^{-r} \\ & =\sum_{r=0}^n{ }^n C_r x^{n-2 r}\end{aligned}$

Hence, the answer is option 1.

Example 4: The coefficient of $x^{10}$ in the expansion of $(1+x)^2\left(1+x^2\right)^3\left(1+x^3\right)^4$ is equal to:

1) 52

2) 56

3) 50

4) 44

Solution:

Expression of Binomial Theorem

$(x+a)^n={ }^n C_0 x^n a^0+{ }^n C_1 x^{n-1} a^1+{ }^n C_2 x^{n-2} a^2+----+{ }^n C_n x^0 a^n$

(when n is a natural number)

The expression is

$
\begin{aligned}
& (1+x)^2\left(1+x^2\right)^3\left(1+x^3\right)^4 \\
& (1+x)^2=x^2+2 x+1 \\
& \left(1+x^2\right)^3=x^6+3 x^4+3 x^2+1 \\
& \left(1+x^3\right)^4=x^{12}+4 x^9+6 x^6+4 x^3+1
\end{aligned}
$
so posible combinations for $x^{10}$ are
$
x \cdot x^9, x \cdot x^3 \cdot x^6, x^2 \cdot x^2 \cdot x^6, x^4 \cdot x^6
$
corresponding coefficients are
$
\begin{aligned}
& 2 \times 4, \quad 2 \times 4 \times 1, \quad 1 \times 3 \times 6, \quad 3 \times 6 \\
& \Rightarrow 8+8+18+18=52
\end{aligned}
$

Hence, the answer is the option 1.

Example 5: The number of terms in the expansion of $(1+x)^{101}\left(1+x^2-x\right)^{100}$ in powers of $x$ is :

1) 302

2) 301

3) 202

4) 101

Solution:

Now,

$\begin{aligned} & (1+x)^{101}\left(1+x^2-x\right)^{100}=(1+x)^{101} \cdot\left(\frac{(1+x)\left(x^2-x+1\right)}{1+x}\right)^{100} \\ & (1+x)^{101} \cdot \frac{\left(x^3+1\right)^{100}}{(1+x)^{100}}=(x+1)\left(x^3+1\right)^{100} \\ &=(1+x)\left(x^3+1\right)^{100} \\ & (1+x)\left({ }^{100} C_0+{ }^{100} C_1 x^3+{ }^{100} C_2 x^6+\ldots .+{ }^{100} C_{100} x^{300}\right) \\ & \Rightarrow \text { No. of terms }=101+101=202\end{aligned}$

Hence, the answer is the option 3.

Summary

Understanding the binomial theorem is essential as it helps to solve the complex equation with powers easily. Binomial coefficients of the term equidistant from the beginning and end are equal. Knowledge of this concept helps to solve and analyze real-life complex problems.