Binomial Theorem for any Index

What is Binomial Expression:

$\begin{aligned}
&\text { An algebraic expression consisting of only two terms is called a Binomial Expression }\\
&e g \cdot(a+b)^2,\left(\sqrt{x}+\frac{k}{x^2}\right)^5,(x+9 y)^{-2 / 3}
\end{aligned}$

Binomial Theorem for any Index

Statement: If $n$ is a rational number and $x$ is a real number such that $|\mathrm{x}|<1$, then,

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

Proof:

Let $f(x)=(1+{{x}})^{{n}}=a_0+a_1 x+a_2 x^2+\ldots+a_1 x^n+\ldots$

$ {{f}(0)}=(1+{{0}} {{{n}}}=1 $

Differentiating (1) w.r.t. $x$ on both sides, we get

$ =a_1+2 a_2 x+3 a_3 x^3+4 a_4 x^3+\ldots+r a_1 x^r-1+\ldots(2) $

Put $x=0$, we get $n=a_1$

Differentiating (2) w.r.t. $\times$ on both sides, we get

$ =2 a_2+6 a_3 x+12 a_4 x^2+\ldots+r(r-1) a_4 x^h-2+\ldots $

Put $x=0$, we get $a_2=[n(n-1)] / 2$ !

Differentiating (3), w.r.t. x on both sides, we get

Put $x=0$, we get $a_3=[n(n-1)(n-2)] / 3$ !

Similarly, we get $a_4=[n(n-1)(n-2)(n-3)] / r!$ and so on

$ \therefore a_n=[n(n-1)(n-2) \ldots(n-r+1)] / r! $

Putting the values of $a_0, a_1, a_2, a_3, \ldots, a_n$ obtained in (1), we get

$ \left(1+x n=1+n x+[\{n(n-1)\} / 2!] x^2+[\{n(n-1)(n-2)\} / 2!] x^3+\ldots+[\{n(n-1)(n-2) \ldots(n-r+\right. $

1) $\} / r!] x^{4}+\ldots$

Hence proved the Binomial theorem of any index.

Results on Binomial Theorem of any Index

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

In the above expansion replace ' $n$ ' with ' $-n$ '

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

$\ldots+\frac{(-n)((-n)-1)((-n)-2) \ldots((-n)-r+1)}{r!} x^r \ldots \ldots \infty $

$\Rightarrow(1+\mathrm{x})^{-\mathrm{n}}=1-\mathrm{nx}+\frac{\mathrm{n}(\mathrm{n}+1)}{2!} \mathrm{x}^2-\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}{3!} \mathrm{x}^3+\ldots \ldots $

$\ldots+(-1)^r \frac{n(n+1)(n+2) \ldots .(n+r-1)}{r!} x^r \ldots \ldots . \infty $

If $-n$ is a negative integer (so that $n$ is a positive integer), then we can re-write this expression as

$ =1-{ }^n C_1 x+{ }^{n+1} C_2 x^2-{ }^{n+2} C_3 x^3+\cdots+{ }^{n+r-1} C_r(-x)^r+\cdots $

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

If $-n$ is a negative integer (so that $n$ is a positive integer), then we can re-write this expression as

$ =1+{ }^n C_1 x+{ }^{n+1} C_2 x^2+{ }^{n+2} C_3 x^3+\cdots+{ }^{n+r-1} C_r(x)^r+\cdots $

Important Note:

The coefficient of $x^5$ in $(1-x)^{-n}$, (when $n$ is a natural number) is ${ }^{n+r-1} C_r$

Some Important Binomial Expansion
1. $(1+x)^{-1}=1-x+x^2-x^3+\cdots$
2. $(1-x)^{-1}=1+x+x^2+x^3+\cdots$
3. $(1+x)^{-2}=1-2 x+3 x^2-4 x^3+\cdots$
4. $(1-x)^{-2}=1+2 x+3 x^2+4 x^3+\cdots$

Recommended Video Based on Binomial Theorem for any Index:

Solved Examples Based on Binomial Theorem for any Index

Example 1: Which of the following Binomial theorem is not possible?
1) $(x+y)^{\frac{-3}{4}}$
2) $\sqrt{(x+y)}$
3) $\sqrt{(x-y)}$
4) $(x+y)^{\frac{7}{3}}$

Solution
As we learnt
Condition for Binomial Theorem for Rational Index:
Here n is a negative integer or a fraction where $-1<n<1$, otherwise expansion will not be possible. for rational powers, we need $-1<n<1$.

Hence, the answer is the option 4.

Example 2: Find the cube root of $318$

1) $6.71$

2) $6.79$

3) $6.83$

4) $6.88$

Solution

The given series is

$
(1+\mathrm{x})^{\mathrm{n}}={ }^{\mathrm{n}} \mathrm{C}_0+{ }^{\mathrm{n}} \mathrm{C}_1 \mathrm{x}+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^2+\ldots .+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}}+\ldots .+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}}
$

for negative or fractional Index

$
(1+x)^n=1+n x+\frac{n(n-1)}{2!} x^2+\frac{n(n-1)(n-2)}{3!} x^3+\ldots \ldots+\frac{n(n-1)(n-2) \ldots . .(n-r+1)}{r!} x^r \ldots \infty
$

Note:
1. If n is negative or fractional index then this condition $|\mathrm{x}|<1$ is essential.
2. There is an infinite number of terms in the expansion of $(1+x)^n$, when $n$ is negative or fractional index.

If the first term is not unity and the index of the binomial is either a negative integer or a fraction, then we expand as follows:

$
\begin{aligned}
(x+a)^n & =\left\{a\left(1+\frac{x}{a}\right)\right\}^n+a^n\left(1+\frac{x}{a}\right)^n \\
& =a^n\left\{1+n \frac{x}{a}+\frac{n(n-1)}{2!}\left(\frac{x}{a}\right)^2+\cdots\right\} \\
& =a^n+n a^{n-1} x+\frac{n(n-1)}{2!} a^{n-2} x^2+\cdots
\end{aligned}
$

The above expansion is valid when $\left|\frac{x}{a}\right|<1$.

$
\begin{gathered}
(318)^{1 / 3}=\left(7^3-25\right)^{1 / 3}=7\left(1-\frac{25}{7^3}\right)^{1 / 3} \\
=7\left(1-\frac{25}{3 \times 343}+\frac{1 \times 2}{3 \times 3 \times 2!}\left(\frac{25}{343}\right)^2+\ldots\right)=7(1-0.0243)=6.83
\end{gathered}
$
Hence, the answer is the option 3.

Example 3: If $0<\mathrm{x}<1$, then the first negative term in the expansion of $(1+x)^{\frac{2 \pi}{5}}$ is
1) $8^{\text {th }}$ term
2) $7^{\text {th }}$ term
3) $6^{\text {th }}$ term
4) $9^{\text {th }}$ term

Solution

Binomial Theorem for any index

For negative or fractional Index and $|x|<1$,

$
(1+x)^n=1+n x+\frac{n(n-1)}{2!} x^2+\frac{n(n-1)(n-2)}{3!} x^3+\ldots \ldots+\frac{n(n-1)(n-2) \ldots \ldots(n-r+1)}{r!} x^r \ldots \infty
$
Now,
The first term is $1$ , so positive
The second term is also positive as both $n$ and $x$ are positive
In any term, $\underline{x}^r$ is positive, and $r!$ is positive. So, the factor that will make a term negative is $(n-r+1)$
So, we need to find $r$ when $(n-r+1)$ will be negative for the first time ( $r$ is an integer), where $n=27 / 5$
Solving $27 / 5-r+1<0$, we get $r>6.4$
So, $r=7$ and this happens in the 8th term, so the 8th term is the answer.
Hence, the answer is the option 1.

Example 4: Find the value of $\sqrt{\frac{y}{x+y}} \cdot \sqrt{\frac{y}{y-x}}$, if x is very small as compared to y .
1) $1-\frac{1}{2} \cdot \frac{y^2}{x^2}$
2) $1+\frac{1}{2} \cdot \frac{y^2}{x^2}$
3) $1-\frac{1}{2} \cdot \frac{x^2}{y^2}$
4) $1+\frac{1}{2} \cdot \frac{x^2}{y^2}$

Solution

Binomial Theorem for any index

For negative or fractional Index and $|x|<1$,

$
(1+x)^n=1+n x+\frac{n(n-1)}{2!} x^2+\frac{n(n-1)(n-2)}{3!} x^3+\ldots \ldots+\ldots \ldots+\frac{n(n-1)(n-2) \ldots . .(n-r+1)}{r!} x^r \ldots \infty$

Now,

$
\begin{aligned}
\sqrt{\frac{y}{x+y}} \sqrt{\frac{y}{y-x}} & =\left(\frac{1}{1+\frac{x}{y}}\right)^{1 / 2}\left(\frac{1}{1-\frac{x}{y}}\right)^{1 / 2} \\
& =\left(1-\frac{x^2}{y^2}\right)^{-1 / 2}=1+\frac{1}{2} \cdot \frac{x^2}{y^2}
\end{aligned}
$

(ignoring higher powers of $x / y$ as $x / y$ is small)
Hence, the answer is the option 4.

For negative or fractional Index and $|x|<1$,

$
(1+x)^n=1+n x+\frac{n(n-1)}{2!} x^2+\frac{n(n-1)(n-2)}{3!} x^3+\ldots \ldots+\frac{n(n-1)(n-2) \ldots . .(n-r+1)}{r!} x^r \ldots \infty
$

Now,

$
\begin{aligned}
\sqrt{\frac{y}{x+y}} \sqrt{\frac{y}{y-x}} & =\left(\frac{1}{1+\frac{x}{y}}\right)^{1 / 2}\left(\frac{1}{1-\frac{x}{y}}\right)^{1 / 2} \\
& =\left(1-\frac{x^2}{y^2}\right)^{-1 / 2}=1+\frac{1}{2} \cdot \frac{x^2}{y^2}
\end{aligned}
$

(ignoring higher powers of $x / y$ as $x / y$ is small)
Hence, the answer is the option 4.

Example 5: To expand $(1+2 x)^{-1 / 2}$ as an infinite series, the range of $x$ should be
1) $\left[-\frac{1}{2}, \frac{1}{2}\right]$
2) $\left(-\frac{1}{2}, \frac{1}{2}\right)$
3) $[-2,2]$
4) $(-2,2)$

Solution
$(1+2 x)^{-1 / 2}$ can be expanded if $|2 x|<1$ i.e., if $|x|<\frac{1}{2}$ i.e., if $\quad-\frac{1}{2}<x<\frac{1}{2}$ i.e., if

$
x \in\left(-\frac{1}{2}, \frac{1}{2}\right)
$
Hence, the answer is the option (2).