Binomial Theorem and its simple applications - Notes, Formula, Examples, Questions

What is Binomial Theorem?

A binomial is an algebraic expression consisting of two distinct terms connected by $+$ or $–$. Understanding binomials is important before diving into the binomial theorem class 11. Let’s compare monomial, binomial, and trinomial:

Examples:

• $x y^2$ (Monomial)

• $x - y, \ y + 4$ (Binomial)

• $x^2 + y + 1$ (Trinomial)

Basic Binomial Identities

Before learning the binomial theorem formula, consider smaller powers of a binomial:

$\begin{aligned} (x+y)^0 &= 1 \\ (x+y)^1 &= x + y \\ (x+y)^2 &= (x+y)(x+y) = x^2 + 2xy + y^2 \\ (x+y)^3 &= (x+y)(x^2 + 2xy + y^2) = x^3 + 3x^2y + 3xy^2 + y^3 \\ (x+y)^4 &= (x+y)(x^3 + 3x^2y + 3xy^2 + y^3) = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4 \end{aligned}$

With smaller powers, it is easy to multiply repeatedly. But what if we need the expansion of $(x+y)^{16}$? This is where the binomial theorem helps!

Observations from Binomial Expansions

From the expansions above, we notice:

1. The total number of terms in the expansion of $(x+y)^n$ is $n+1$. For example, $(x+y)^3$ has $4$ terms.

2. The powers of $x$ start from $n$ and decrease to $0$, while powers of $y$ start from $0$ and increase to $n$.

3. In every term, the sum of the indices of $x$ and $y$ equals $n$, the power of the binomial.

4. The coefficients in the expansion are called binomial coefficients, denoted as $\binom{n}{r} = C(n, r) = { }^n C_r = \frac{n!}{r!(n-r)!}$

5. The Binomial coefficient are symmetric, i.e., $\binom{n}{r} = \binom{n}{n-r}$.

These principles form the foundation of binomial theorem class 11 formulas, useful for binomial theorem examples, proofs, and miscellaneous exercises.

These patterns lead us to the Binomial Theorem, which can be used to expand any binomial.

Binomial Theorem states that,

If $n$ is any positive integer, then

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

where $\binom{n}{r}=C(n, r)={ }^n C_r=\frac{n!}{r!(n-r)!}$.

Binomial Theorem Formula

The general formula of binomial theorem to expand binomial expressions with higher power is

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

where $\binom{n}{r}=C(n, r)={ }^n C_r=\frac{n!}{r!(n-r)!}$.

Binomial Theorem Proof

Statement:

If $n$ is any positive integer, then

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

where $\binom{n}{r}=C(n, r)={ }^n C_r=\frac{n!}{r!(n-r)!}$.

Proof:

We can prove the binomial theorem using mathematical induction on $n$.
Base Case $(n=1)$ :
For $n=1$, the binomial expansion of $(x+y)^1$ is:

$
(x+y)^1=x+y
$
On the other hand, the binomial formula gives:

$
(x+y)^1=\displaystyle\sum_{k=0}^1\binom{1}{k} x^{1-k} y^k=\binom{1}{0} x^1 y^0+\binom{1}{1} x^0 y^1=x+y
$

So, the base case holds.
Inductive Hypothesis:
Assume that the binomial theorem holds for some integer $n=m$. That is, assume:

$
(x+y)^m=\displaystyle\sum_{k=0}^m\binom{m}{k} x^{m-k} y^k
$
Inductive Step:
We want to show that the binomial theorem holds for $n=m+1$, i.e.,

$
(x+y)^{m+1}=\displaystyle\sum_{k=0}^{m+1}\binom{m+1}{k} x^{(m+1)-k} y^k
$

To do this, consider:

$
(x+y)^{m+1}=(x+y)(x+y)^m
$

Using the inductive hypothesis, we expand $(x+y)^m$ :

$
(x+y)^{m+1}=(x+y) \displaystyle\sum_{k=0}^m\binom{m}{k} x^{m-k} y^k
$

Now distribute $(x+y)$ over the sum:

$
=\displaystyle\sum_{k=0}^m\binom{m}{k} x^{m+1-k} y^k+\displaystyle\sum_{k=0}^m\binom{m}{k} x^{m-k} y^{k+1}
$

We can now combine these two sums into a single sum. Observe that in the first sum, the powers of $x$ start at $m+1$ and decrease, while in the second sum, the powers of $y$ increase starting from 1.
Reindex the second sum by letting $j=k+1$, so that:

$
=\displaystyle\sum_{k=0}^m\binom{m}{k} x^{m+1-k} y^k+\displaystyle\sum_{j=1}^{m+1}\binom{m}{j-1} x^{m+1-j} y^j
$
Now, combine the terms where the powers of $x$ and $y$ match:

$
=x^{m+1}+\displaystyle\sum_{k=1}^m\left(\binom{m}{k}+\binom{m}{k-1}\right) x^{m+1-k} y^k+y^{m+1}
$
Finally, use the identity $\binom{m+1}{k}=\binom{m}{k}+\binom{m}{k-1}$ to rewrite the sum:

$
=\displaystyle\sum_{k=0}^{m+1}\binom{m+1}{k} x^{(m+1)-k} y^k
$
This completes the inductive step.

Thus, by mathematical induction, the binomial theorem holds for all $n \geq 1$.

Properties of Binomial Theorem

Some important properties of binomial theorem are

• The total number of each and every term in the expansion $(x+y)^n$ is $n + 1$ .
• The sum total of the indices of $x$ and $y$ in each term is $n$.
• The expansion shown above is also true when both $x$ and $y$ are complex numbers.
• The coefficient of all the terms is equidistant (equal in distance from each other) from the beginning to the end.
• The values of these binomial coefficients gradually go up to the maximum and progressively lessen.

Terms in Binomial Theorem

General Term in Binomial Expansion

Expanding a binomial with a high exponent such as $(x+2 y)^{16}$ can be a lengthy process. Sometimes we may be interested only in a certain term of a binomial expansion. To find this specific term, we do not need to completely expand the binomial. Now, let us note the pattern of coefficients in the expansion of $(x+y)^5$

$
(x+y)^5=x^5+\binom{5}{1} x^4 y+\binom{5}{2} x^3 y^2+\binom{5}{3} x^2 y^3+\binom{5}{4} x y^4+y^5
$

The second term is
$\binom{5}{1} x^4 y$
The third term is
In this way, we can generalize this result.

The $(r+1)$ th term of the binomial expansion of $(x+y)^n$ is:

$
\binom{n}{r} \mathrm{x}^{\mathrm{n}-\mathrm{r}} \mathrm{y}^{\mathrm{r}}
$

Middle term(s) of Binomial Theorem

The middle term in the expansion $(x+y)^n$, depends on the value of ' $n$ '

Case 1: When ' $n$ ' is even
If n is even, and the number of terms in the expansion is $\mathrm{n}+1$, so $\mathrm{n}+1$ is odd number therefore only one middle term is obtained which is

$
\left(\frac{\mathrm{n}}{2}+1\right)_{\text {term. }}^{\text {th }}
$

It is given by

$
\mathrm{T}_{\frac{n}{2}+1}=\binom{n}{\frac{n}{2}} x^{\frac{n}{2}} y^{\frac{n}{2}}
$
Case 2: When ' $n$ ' is odd
In this case, the number of terms in the expansion will be $\mathrm{n}+1$. Since n is odd so, $\mathrm{n}+1$ is even. Therefore, there will be two middle terms in the expansion, namely $\left(\frac{n+1}{2}\right)^{t h}\left(\frac{n+3}{2}\right)^{t h}$ and terms. It is given by

$
T_{\frac{n+1}{2}}=\binom{n}{\frac{n-1}{2}} x^{\frac{n+1}{2}} \cdot y^{\frac{n-1}{2}} \text { and } \quad T_{\frac{n+3}{2}}=\binom{n}{\frac{n+1}{2}} x^{\frac{n-1}{2}} \cdot y^{\frac{n+1}{2}}
$

Independent Term

Independent term is the constant term in the expansion of the binomial expansion. In the expansion of $\left(x+\frac{1}{x}\right)^{2 n}$, where $x \neq 0$, the middle term is $\left(\frac{2 n+1+1}{2}\right)^{t h}$, i.e., $(n+1)^{\mathrm{th}}$ term, as $2 n$ is even.

It is given by ${ }^{2 n} \mathrm{C}_n x^n\left(\frac{1}{x}\right)^n={ }^{2 n} \mathrm{C}_n$ (constant).
This term is called the term independent of $x$ or the constant term.

Numerically Greatest Value

The numerical value of each term of the binomial expansion is determined by the value of the Binomial coefficients. Numerically greatest value is defined as the largest term among the product of the variable coefficients(Binomial coefficients) in the Binomial expansion. In general, Numerically greatest value of the Binomial expansion of $(x+a)^n$ is the $r$th and $(r+1)$th term where $r=\frac{(n+1)}{1+|\frac{x}{a}|}$. It is represented as $T_{r}$ and $T_{r+1}$.

Ratio of Consecutive Terms

The ratio of two consecutive terms in a binomial expansion of the form $(a+b)^n$ is $\frac{n-r+1}{r}$

Application of Binomial Theorem

From calculating probabilities, expanding algebraic expressions quickly, to solving problems in combinatorics, the theorem simplifies complex calculations. In this section, you will explore real-life applications of binomial theorem, its use in class 11 problems, and how it helps in expanding expressions efficiently.

Finding Remainder Using Binomial Theorem

To find the remainder of an expression using Binomial theorem,

If the number is given in the form of ' $a^n$ ' and which is divided by ' $b$ '. To find the remainder, adjust the power of ' $a$ ' to $a^m$ such that it is very close to ' $b$ ' with a difference of 1 (i.e. $b+1$ or $b-1$).

Also, when number of the type $7 \mathrm{k}-1$ is divided by $7$ , remainder cannot be -1, as remainder is always positive

So, in such cases, we have $\frac{7 k-1}{7}=\frac{7 k-7+6}{7}=k-1+\frac{6}{7}$

Hence, the reminder is $6(=7-1)$

For example,

If $32^{30}$ is divided by 7, then the remainder is

We know that $32=2^5$, so, $32^{30}$ can be written as

$\left(2^5\right)^{{30}} =2^{150}=\left(2^3\right)^{50}=8^{50}=(7+1)^{50} $

$ =\left[(7)^{50}+{ }^{50} \mathrm{C}_1(7) {{49}}={ }^{50} \mathrm{C}_2(7) {{48}+\ldots+1]}\right. $

$ =7\left[(7)^{49}+{ }^{50} \mathrm{C}_1(7) {{48}}+{ }^{50} \mathrm{C}_2(7) {{47+}}+1\right.$

$ =7 \mathrm{k}+1$

$ \Rightarrow \text { remainder is } 1 $

Last digits of an expression

To find the last digits of an expansion using Binomial theorem,

If the given expression is $a^n$ then write the expression in the form of $(10 k \pm 1)^m$, where k and m are positive integers

Now take $10$ common for getting last digit, $100$ for getting last two digit, $1000$ for getting last $3$ digits and so on ...

After expanding $\left(10 \mathrm{k} \pm 1{ }^{m}\right.)$ using binomial theorem, it will look like

$(10 \mathrm{k} \pm 1)^{\mathrm{m}}= (10 \mathrm{k})^{\mathrm{m}}+{ }^{\mathrm{m}} \mathrm{C}_1(10 \mathrm{k})^{\mathrm{m}-1}( \pm 1)+{ }^{\mathrm{m}} \mathrm{C}_2(10 \mathrm{k})^{\mathrm{m}-2}( \pm 1)^2 $

$+{ }^{\mathrm{m}} \mathrm{C}_3(10 \mathrm{k})^{\mathrm{m}-2}( \pm 1)^3 \ldots \ldots \ldots+{ }^{\mathrm{m}} \mathrm{C}_{\mathrm{m}-1}(10 \mathrm{k})( \pm 1)^{\mathrm{m}-1}+( \pm 1)^{\mathrm{m}} $

Hence, the number is $10 \alpha+( \pm 1)^m$ (this last part decides the last digit)

The number can also be written as $1008+{ }^m C_{m-1}(10 k)( \pm 1)^{m-1}+( \pm 1)^m$ (last 2 terms decide the last 2 digits) $\alpha, \beta$ are Integers

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.

Real-world Applications of Binomial Theorem

The application of binomial theorem in real life are

1. The binomial theorem is used heavily in Statistical and Probability Analyses. It is so much useful as our economy depends on Statistical and Probability Analyses.

2. In higher mathematics and calculation, the Binomial Theorem is used in finding roots of equations in higher powers. Also, it is used in proving many important equations in physics and mathematics.

3. In Weather Forecast Services,

4. Ranking up candidates

5. Architecture, estimating cost in engineering projects.

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