1 Million Means: 1 million in numerical is represented as 10,00,000. The Indian equivalent of a million is ten lakh rupees. It is not a...

1 Million Means: 1 Million in Rupees, Lakhs and Crores

June 5, 2024**Binomial Theorem**: The binomial theorem is the most commonly used theorem in mathematics. The binomial theorem is a technique for expanding a binomial expression raised to any finite power. It is used to solve problems in combinatorics, algebra, calculus, probability etc. It is used to compare two large numbers, to find the remainder when a number raised to some large exponent is divided by another number and used in the probability to find the success or failure of an experiment. The binomial theorem is also used in weather forecasting, predicting the national economy in the next few years and distribution of IP addresses. Let’s learn about the Binomial theorem in detail.

The Binomial Theorem is a technique for expanding a binomial expression raised to any finite power. An algebraic expression with two distinct terms is known as a binomial expression. For example \(a + b,\;\,2x – {y^3}\) etc. The algebraic expansion of binomial powers is described by the binomial theorem, which use Pascal’s triangles to calculate coefficients.

The Binomial Theorem states that for a non-negative integer \(n,\)

\({(a + b)^n} = \mathop \sum \limits_{r = 0}^n \left( {\begin{array}{*{20}{l}}n\\r\end{array}} \right){a^{n – r}}{b^r}\)

where \(\left( {\begin{array}{*{20}{c}} n\\ r \end{array}} \right) = {}^n{C_r} = \frac{{n!}}{{r!\left( {n – r} \right)!}}\) is binomial coefficient.

we can use the pascal triangle to determine the coefficient of binomial expansion like

Just look at the coefficients in the expressions above; we will find a pattern like this as the exponent increases.

Similarly, all the coefficient of the binomial expansion is identified. In other words, the coefficients of binomial expansion are the same as the entries in the \({n^{{\rm{th}}}}\) row of Pascal’s Triangle.

The expansion of \({\left( {a + b} \right)^n}\) is given by

\({\left( {a + b} \right)^n} = {\;^n}{C_0}{a^n}{b^0} + {\;^n}{C_1}{a^{n – 1}}{b^1}

+ {\;^n}{C_2}{a^{n – 2}}{b^2} + \ldots + {\;^n}{C_n}{a^0}{b^n}\)

\({\left( {a + b} \right)^n} = \mathop \sum \limits_{r = 0}^n {\;^n}{C_r}{a^{n – r}}{b^r}\) where \({\;^n}{C_r} = \frac{{n!}}{{\left( {n – r} \right)!r!}}\) and \(n \in N,r \in W,r \le n\)

The general term of the binomial expansion of \({\left( {a + b} \right)^n}\) is given by

\({T_{r + 1}} = {\;^n}{C_r}{a^{n – r}}{b^r}\)

So,

when \(r = 0 \Rightarrow {T_1}\)

\(r = 1 \Rightarrow {T_2}\) and so on.

An important point to note :

1. The total number of terms in the expansion of \({\left( {a + b} \right)^n}\) are \(n + 1\)

2. The sum of powers of \(a\) and \(b\) always \(n\)

3. \({\;^n}{C_0},{\;^n}{C_1},\;{\;^n}{C_2}, \ldots ,{\;^n}{C_n}\) are binomial coefficients

If someone asks you to find the \({4^{{\rm{th}}}}\) term in the expansion of \({\left( {2 + 3x} \right)^7},\) you can easily say \({T_4} = {\;^7}{C_3}{a^{7 – 3}}{b^3} = {\;^7}{C_3}{\left( 2 \right)^{7 – 3}}{\left( {3x} \right)^3} = 35 \times 16 \times 27 \times {x^3} = 15120\;{x^3}\)

Using binomial theorem, we can also find the middle term.

The middle term of binomial expansion of \({\left( {a + b} \right)^n}\) depends upon the value of \(n\)

(1) When \(n\) is even, then total number of terms in the expansion of \({(a + b)^n}\) is \(n + 1\) (odd).<
>So, \({\left( {\frac{n}{2} + 1} \right)^{{\rm{th\;}}}}\) term is only middle term. \(T\left( {\frac{n}{2} + 1} \right) = {}^n{C_{\frac{n}{2}}}{a^{\frac{n}{2}}}{b^{\frac{n}{2}}}\)

(2) When \(n\) is odd, then total number of terms in the expansion of \({(a + b)^n}\) is \(n + 1\) (even).

So, \({\left( {\frac{{n + 1}}{2}} \right)^{{\rm{th\;}}}}\) and \({\left( {\frac{{n + 3}}{2}} \right)^{{\rm{th\;}}}}\) are two middle terms.

\({{\rm{T}}_{\left( {\frac{{n + 1}}{2}} \right)}} = {\;^n}{{\rm{C}}_{\frac{{n – 1}}{2}}}\;{a^{\frac{{n + 1}}{2}}}\;{b^{\frac{{n – 1}}{2}}}\) and \({T_{\left( {\frac{{n + 3}}{2}} \right)}} = {\;^n}{C_{\frac{{n + 1}}{2}}}\;{a^{\frac{{n – 1}}{2}}}\;{b^{\frac{{n + 1}}{2}}}\)

Some of the properties of binomial theorem which helps to solve a lot of problems are as follows:

For the sake of convenience, the coefficients \(^n{C_0},{\;^n}{C_1}, \ldots .{\;^n}{C_n}\) are usually represented as \({C_0},\,{C_1},\,….,\,{C_{n.}}\)

- \({C_0} + {C_1} + {C_2} + \ldots + {C_n} = {2^n}\)
- \({C_0} – {C_1} + {C_2} – \ldots + {( – 1)^n}{C_n} = 0\)
- \({C_0} + {C_2} + {C_4} + \ldots = {C_1} + {C_3} + {C_5} + \ldots = {2^{n – 1}}\)
- \(^n{C_r} = {\;^n}{C_{n – r}}\)
- \(r\left( {{}^n{C_r}} \right) = n\,{}^{n – 1}{C_{r – 1}}\)
- \(\frac{{{\;^n}{C_r}}}{{r + 1}} = \frac{{{\;^{n + 1}}{C_{r + 1}}}}{{n + 1}}\)
- \(^n{C_r} + {\;^n}{C_{r – 1}} = {\;^{n + 1}}{C_r}\)

Where \(n \in N,\,r \in W\) and \(r \le n\)

Some examples are given below is solved with the concept of the binomial theorem.

Example: What will be remainder when \({7^{103}}\) is divided by \(25\)

Solution:

\({7^{103}} = 7{\left( {50 – 1} \right)^{51}} = 7\left( {{{50}^{51}} – {}^{51}{C_1}{{50}^{50}} + {}^{51}{C_2}{{50}^{49}} – …. – 1} \right)\)

\(= 7\left( {{{50}^{51}} – {}^{51}{C_1}{{50}^{50}} + … + {}^{51}{C_{50}}50} \right) – 7 – 18 + 18\)

\(= 7\left( {{{50}^{51}} – {}^{51}{C_1}{{50}^{50}} + … + {}^{51}{C_{50}}50} \right) – 25 + 18\)

\( \Rightarrow \) remainder is \(18.\)

Example: What will be the last two digits of \({11^{25}}\) ?

Soluion:

\({11^{25}} = {\left( {1 + 10} \right)^{25}} = \left( {\begin{array}{*{20}{c}}{25}\\0\end{array}} \right) + 10 \cdot \left( {\begin{array}{*{20}{c}}{25}\\1\end{array}} \right) + \sum\limits_{k = 2}^{25} {\left( {\begin{array}{*{20}{c}}{25}\\k\end{array}} \right)} \cdot {10^k}\)

\(= 1 + 25 \cdot 10 + \sum\limits_{k = 2}^{25} {\left( {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{25}\\k\end{array}}\end{array}} \right)} \cdot {10^k}\)

\( = 1 + 250 + {10^2}.\left( {\sum\limits_{k = 2}^{25} {\left( {\begin{array}{*{20}{c}}
{25}\\
k
\end{array}} \right)} {{.10}^{k – 2}}} \right)\)

Hence, the last two digits are \(5\) and \(1\)

Example: If \(n\) is a positive integer, prove that \({3^{3n}} – 26n – 1\) is divisible by \(676\)

Solution:

\(\left( {{3^{3n}} – {{26}^n} – 1} \right) = {\left( {27} \right)^n} – 26n – 1\)

\(= {\left( {1 + 26} \right)^n} – 26n – 1\)

\( = 1 + 26n + {\left( {26} \right)^2}\left[ {{}^n{C_2} + 26\,{}^n{C_3} + – – – + {{\left( {26} \right)}^{n – 2}}} \right] – 26n – 1\)

Let \(\left[ {{\;^n}{C_2} + {{26}^n}{C_3} + – – – + {{\left( {26} \right)}^{n – 2}}} \right] = k\)

\(= 676\,{\rm{k}}\)

\(\therefore \left( {{3^{3n}} – 26n – 1} \right)\) is divisible by \(676\)

Example: Which is larger \({\left( {1.1} \right)^{10000}}\;\) or \(1000\)?

Solution:

\({\left( {1.1} \right)^{10000}} = {\left( {1 + 0.1} \right)^{10000}}\)

\(= \left[ {{\;^{10000}}{C_0}{{\left( 1 \right)}^{10000}}{{\left( {0.1} \right)}^0} + {\;^{10000}}{C_1}{{\left( 1 \right)}^{9999}}{{\left( {0.1} \right)}^1} + {\;^{10000}}{C_2}{{\left( 1 \right)}^{9998}}{{\left( {0.1} \right)}^2} + \ldots + {\;^{10000}}{C_{10000}}{{\left( 1 \right)}^0}{{\left( {0.1} \right)}^{10000}}} \right]\)

\( = \left[ {1 + 10000 \times \left( {0.1} \right) + {\;^{10000}}{{\rm{C}}_2}{{\left( {0.1} \right)}^2} + – – } \right]\)

\( = 1 + 1000 + \) (Some positive term)

\(= 1001 + \) (Some positive term)

So, clearly \({\left( {1.1} \right)^{10000}} > 1000\)

*Q.1. Evaluate the following expression \({\left( {1 + 2\sqrt x \;} \right)^5} + {\left( {1 – 2\sqrt x \;} \right)^5}\)*** Ans**: \({\left( {1 + 2\sqrt x } \right)^5} + {\left( {1 – 2\sqrt x } \right)^5}\)

\(= \left[ {{\;^5}{C_0} + {\;^5}{C_1}\left( {2\sqrt x } \right) + {\;^5}{C_2}{{\left( {2\sqrt x } \right)}^2} + – – -+ {\;^5}{C_5}{{\left( {2\sqrt x } \right)}^5}} \right]\)

\(+ \left[ {{\;^5}{C_0} – {\;^5}{C_1}\left( {2\sqrt x } \right) + {\;^5}{C_2}{{\left( {2\sqrt x } \right)}^2} + – – – {\;^5}{C_5}{{\left( {2\sqrt x } \right)}^5}} \right]\)

\(= 2\left[ {{\;^5}{C_0} + {\;^5}{C_2}{{\left( {2\sqrt x } \right)}^2} + {\;^5}{C_4}{{\left( {2\sqrt x } \right)}^4}} \right]\)

\(= 2\left[ {1 + 10 \times 4 \times x + 5 \times 16{x^2}} \right]\)

\(= 160{x^2} + 80x + 2\)

** Q.2. Using the binomial theorem, prove that \({2^{3n}} – 7n – 1\) is divisible by \(49\) where \(n \in N.\)**: \(\left( {{2^{3n}} – 7n – 1} \right) = {\left( {{2^3}} \right)^n} – 7n – 1\)

Ans

\(= {8^n} – 7n – 1\)

\(= {\left( {1 + 7} \right)^n} – 7n – 1\)

\(= \left[ {{\;^n}{C_0}{7^0} + {\;^n}{C_1}{7^1} + {\;^n}{C_2}{7^2} + – – – + {\;^n}{C_n}{7^n}} \right] – 7n – 1\)

\(= \left[ {1 + 7n + {\;^n}{C_2} \cdot {{\left( 7 \right)}^2} + {\;^n}{C_3} \cdot {{\left( 7 \right)}^3} + – – – + {\;^n}{C_{n – 1}} \cdot {{\left( 7 \right)}^{n – 1}} + {\;^n}{C_n} \cdot {{\left( 7 \right)}^n}} \right] – 7n – 1\)

\(= {7^2}\left[ {{\;^n}{C_2} + {\;^n}{C_3}.7 + – – – + {\;^n}{C_{n – 1}} \cdot {{\left( 7 \right)}^{n – 3}} + {\;^n}{C_n} \cdot {{\left( 7 \right)}^{n – 2}}} \right]\)

\(= 49\left[ {{\;^n}{C_2} + {\;^n}{C_3} \cdot 7 + – – + n.{{\left( 7 \right)}^{n – 3}} + {{\left( 7 \right)}^{n – 2}}} \right]\)

Put \(k = \left[ {{\;^n}{C_2} + {\;^n}{C_3} \cdot 7 + – – + n \cdot {{\left( 7 \right)}^{n – 3}} + {{\left( 7 \right)}^{n – 2}}} \right]\)

\(= 49\,{\rm{k}}\)

\(\therefore \left( {{2^{3n}} – 7n – 1} \right)\) is divisible by \(49.\)

*Q.3. If the \({7^{{\rm{th}}}}\) term in the expansion of \({\left( {3{x^2} – \frac{1}{{{x^3}}}} \right)^3}\) is \(\frac{k}{{{x^{10}}}},\) then find the value of \(k\)*** Ans**: We know that \({(r + 1)^{{\rm{th}}}}\) term in the expansion of \({\left( {x + a} \right)^n}\) is \({T_{r + 1}} = {\;^n}{C_r}{x^{n – r}}{a^r}.\)

\(\therefore {7^{{\rm{th\;}}}}\) term in expansion of \({\left( {3{x^2} – \frac{1}{{{x^3}}}} \right)^{10}}\) is \({T_7} = {T_{6 + 1}} = {\;^{10}}{C_6}{\left( {3{x^2}} \right)^{10 – 6}} \times {\left( {\frac{{ – 1}}{{{x^3}}}} \right)^6}\)

\(\Rightarrow {T_7} = {\;^{10}}{C_6}\frac{{{3^4}{x^8}}}{{{x^{18}}}}\)

\(\Rightarrow {T_7} = \frac{{10!}}{{6! \times 4!}} \times \frac{{{3^4}}}{{{x^{10}}}}\)

\(\Rightarrow {T_7} = 210 \times \frac{{81}}{{{x^{10}}}}\)

\(\Rightarrow {T_7} = \frac{{17010}}{{{x^{10}}}}\)

It is given that, \(\frac{k}{{{x^{10}}}} = \frac{{17010}}{{{x^{10}}}}\)

\(\Rightarrow k = 17010\)

** Q.4. Find the \({4^{{\rm{th}}}}\) term from the beginning and \({4^{{\rm{th}}}}\) term from the end in the expansion of \({\left( {x + \frac{2}{x}} \right)^9}\)**: Given: \({\left( {x + \frac{2}{x}} \right)^9}\)

Ans

Let \({\left( {r + 1} \right)^{{\rm{th}}}}\) be the \({4^{{\rm{th}}}}\) term from the end

Then, \({T_{r + 1}}\) is \(\left( {10 – 4 + 1} \right){\;^{{\rm{th}}}}\) i.e., \({7^{{\rm{th}}}}\) term from the beginning.

\({T_7} = {T_{6 + 1}}\)

\( = {\;^9}{C_6}\left( {{x^{9 – 6}}} \right){\left( {\frac{2}{x}} \right)^6}\)

\(= \frac{{9 \times 8 \times 7}}{{3 \times 2}}\left( {{x^3}} \right)\left( {\frac{{64}}{{{x^6}}}} \right)\)

\(= \frac{{5376}}{{{x^3}}}\)

\({4^{{\rm{th}}}}\) term from the beginning \(= {T_4} = {T_{3 + 1}}\)

\({T_4} = {\;^9}{C_3}\left( {{x^{9 – 3}}} \right){\left( {\frac{2}{x}} \right)^3}\)

\(= \frac{{9 \times 8 \times 7}}{{3 \times 2}}\left( {{x^6}} \right)\left( {\frac{8}{{{x^3}}}} \right)\)

\( = 672{{\rm{x}}^3}\)

** Q.5. Find the coefficient of \(x\) Find the coefficient of \(\left( {1 – 2{x^3} + 3{x^5}} \right){\left( {1 + \frac{1}{x}} \right)^8}\)**: \(\left( {1 – 2{x^3} + 3{x^5}} \right){\left( {1 + \frac{1}{x}} \right)^8}\)

Ans

\(= \left( {1 – 2{x^3}} \right.\left. { + 3{x^5}} \right)\left( {{\;^8}{C_0} + {\;^8}{C_1}\frac{1}{x} + {\;^8}{C_2}\frac{1}{{{x^2}}} + {\;^8}{C_3}\frac{1}{{{x^3}}} + {\;^8}{C_4}\frac{1}{{{x^4}}} + \ldots \ldots + {\;^8}{C_8}\frac{1}{{{x^8}}}} \right)\)

In order to find the coefficient of term containing \(x\) in the binomial expansion of

\(\left( {1 – 2{x^3} + 3{x^5}} \right){\left( {1 + \frac{1}{x}} \right)^8}.\)

We need such terms from the expansion of \({\left( {1 + \frac{1}{x}} \right)^8}\) which give resultant terms containing \(x\) on multiplying with \(\left( {1 – 2{x^3} + 3{x^5}} \right).\)

As we can see terms of binomial expansion of \({\left( {1 + \frac{1}{x}} \right)^8}\) are having only negative powers.

So, the terms in the binomial expansion of \({\left( {1 + \frac{1}{x}} \right)^8}\) contain \({x^{ – 2}}\) and \({x^{ – 4}}\) are desired terms.

Let \({\left( {r + 1} \right)^{{\rm{th}}}}\) term contains \({x^{ – 2}}\) and \({\left( {p + 1} \right)^{{\rm{th}}}}\) term contains \({x^{ – 4}}\) in an expansion of \({\left( {x + \frac{1}{x}} \right)^8}.\)

\(\therefore {T_{r + 1}} = {\;^8}{C_r}{\left( 1 \right)^{8 – {\rm{r}}}} \cdot {\left( {\frac{1}{x}} \right)^{\rm{r}}} = {\;^8}{C_{\rm{r}}} \cdot {\left( x \right)^{ – r}}\) and \({T_{p + 1}} = {\;^8}{C_p}{\left( 1 \right)^{8 – p}} \cdot {\left( {\frac{1}{x}} \right)^p} = {\;^8}{C_r} \cdot {\left( x \right)^{ – p}}{\rm{\;}} \ldots \)

\({\left( {r + 1} \right)^{{\rm{th}}}},{\left( {p + 1} \right)^{{\rm{th}}}}\) terms contain \({x^{ – 2}}\) and \({x^{ – 4}}\) respectively.

\(\therefore {\rm{\;}} – {\rm{r}} = – 2 \Rightarrow r = 2\) and \(- p = – 4 \Rightarrow p = 4\)

Put \(r = 2\) and \(p = 4\) in (1) we get,

\({T_{r + 1}} = {\;^8}{C_2}{x^{ – 2}}\) and \({T_{p + 1}} = {\;^8}{C_4}{x^{ – 4}}\)

So, and \(\left( {3{x^5}} \right) \cdot {\;^8}{C_4}{x^{ – 4}} = 3 \times {\;^8}{C_4} \cdot x\)

\(\therefore\) The term contains \(x\) in expansion \(\left( {1 – 2{x^3} + 3{x^5}} \right){\left( {1 + \frac{1}{x}} \right)^8}\) is \(x\left( { – 2 \times {\;^8}{C_2} + 3 \times {\;^8}{C_4}} \right)\) and the coefficient of term contains \(x\) in the expansion \(\left( {1 – 2{x^3} + 3{x^5}} \right){\left( {1 + \frac{1}{x}} \right)^8}\) is

\(\left( { – 2 \times {\;^8}{C_2} + 3 \times {\;^8}{C_4}} \right) \Rightarrow – 2 \times 28 + 3 \times 70 = \, – 56 + 210 \Rightarrow 154\)

Special cases of the binomial theorem have been known since at least the 4th century BCE when the Greek mathematician Euclid mentioned the special case of the binomial theorem for exponent 2. There is evidence that the binomial theorem for the cube was known in India by the 6th century AD.

Binomial coefficients, as combinatorial quantities expressing the number of ways of choosing k objects out of n without replacement, were of interest to ancient Indian mathematicians. The earliest known reference to this belligerent problem is the Chandashastra of the Indian lyricist Pingala (200 BCE), which contains a method for its solution. The 10th century AD commentator Halayudh explains this method using the method now known as Pascal’s triangle.

By the 6th century AD, Indian mathematicians probably knew how to express it as a quotient (\(\frac{n!}{\displaystyle(n-k)!k!}\)) and a clear statement of this rule can be found in the 12th-century text Lilavati by Bhaskara. The first formulation of the binomial theorem and the table of binomial coefficients can, to our knowledge, be found in a work by al-Qaraji, which is cited by al-Samawal in his “Al-Bahir”. Al-Qaraji described the triangular pattern of binomial coefficients and also provided mathematical proofs of both the binomial theorem and Pascal’s triangle, using an early form of mathematical induction. The Persian poet and mathematician Omar Khayyam was probably familiar with the high order formula, although many of his mathematical works have disappeared.

The binomial expansion of smaller degrees was known in the 13th-century mathematical works of Yang Hui and Chu Shih-Chih. Yang Hui attributes this method to a much earlier 11th-century text by Jia Jian, although those writings are still lost. In 1544, Michael Stifel introduced the term “binomial coefficients” and showed how to use them to express according to \((1+a)^n(1+a)^{n-1}\), via “Pascal’s Triangle”. Blaise Pascal in his “Trate des Triangle” Arithmetic made extensive study of symmetry triangles. However, the pattern of numbers was already known to European mathematicians of the late Renaissance, including Stifel, Niccol Fontana Tartaglia, and Simon Stevin.

Isaac Newton is generally credited with the generalized binomial theorem, which is valid for any rational exponent.

- The binomial theorem is mentioned in the song Major-Gen in the comic opera The Pirates of Penzance.
- Professor Moriarty is described by Sherlock Holmes as writing a treatise on the binomial theorem.
- The Portuguese poet Fernando Pessoa, using the odd name of lvaro de Campos, wrote that “Newton’s binomial is as beautiful as Venus de Milo. The truth is that very few people notice it.”
- In the 2014 film The Imitation Game, Alan Turing references Isaac Newton’s work on the binomial theorem during his first meeting with Commander Denniston at Bletchley Park.

For positive values of a and b, with the binomial theorem n = 2 the geometrically obvious fact that is a square of side a + b can be cut into sides of a square a, side of a square b, and with sides Two rectangles a and b. With n = 3, the theorem states that a cube of side a + b can be cut into a cube of side a, a cube of side b, three a × a × b rectangular boxes, and three a × b × b rectangular box.

In this article, we have learnt about the Binomial theorem, its properties and its examples and got to know that it plays an essential role in the mathematical world. In mathematics, It is used to solve problems in combinatorics, algebra, calculus, probability, statistics and also in data science. Binomial coefficients, as combinatorial quantities expressing the number of ways of choosing k objects out of n without replacement, were of interest to ancient Indian mathematicians.

* Q.1. How to find the middle term in binomial theorem?*: The middle term of binomial expansion of \({\left( {x + y} \right)^n}\) depends upon the value of \(n\)

Ans

(1) When \(n = \) even, then the total number of terms in the expansion of \({\left( {x + y} \right)^n}\) is \(n + 1\)(odd).

So, \({\left( {\frac{n}{2} + 1} \right)^{{\rm{th\;}}}}\) term is only middle term. \({T_{\left( {\frac{n}{2} + 1} \right)}} = {}^n{C_{\frac{n}{2}}}{x^{\frac{n}{2}}}{y^{\frac{n}{2}}}\)

(2) When \(n = \) odd, then total number of terms in the expansion of \({\left( {x + y} \right)^n}\) is \(n + 1\) (even)

So, \({\left( {\frac{{n + 1}}{2}} \right)^{{\rm{th\;}}}}\) and \({\left( {\frac{{n + 3}}{2}} \right)^{{\rm{th\;}}}}\) are two middle terms.

\({T_{\left( {\frac{{n + 1}}{2}} \right)}} = {\;^n}{C_{\frac{{n – 1}}{2}}}\;{x^{\frac{{n + 1}}{2}}}\;{y^{\frac{{n – 1}}{2}}}\) and \({T_{\left( {\frac{{n + 3}}{2}} \right)}} = {\;^n}{C_{\frac{{n + 1}}{2}}}\;{x^{\frac{{n – 1}}{2}}}\;{y^{\frac{{n + 1}}{2}}}\)

** Q.2. Who invented the binomial theorem?**: Isaac Newton discovered binomial theorem in \(1665\) and later stated in \(1676\) without proof but the general form and its proof for any real number \(n\) was published by John Colson in \(1736.\)

Ans

** Q.3. State binomial theorem.**: The Binomial Theorem states that for a non-negative integer \(n,\)

Ans

\({\left( {a + b} \right)^n} = \mathop \sum \limits_{k = 0}^n \left( {\begin{array}{*{20}{l}}n\\k\end{array}} \right){a^{n – k}}{b^k}\)

where \(\left( {\begin{array}{*{20}{l}}n\\k\end{array}} \right) = \frac{{n!}}{{k!\left( {n – k} \right)!}}\) is a binomial coefficient. In other words, the coefficients when \({\left( {a + b} \right)^n}\) is expanded and like terms are collected are the same as the entries in the \(n\) th row of Pascal’s Triangle. The total number of terms in the expansion of \({\left( {a + b} \right)^n}\) and \(n + 1.\) The sum of powers of \(a\) and \(b\) is always \(n.\) \({\;^n}{C_0},{\;^n}{C_1},\;{\;^n}{C_2}, \ldots ,{\;^n}{C_n}\) are binomial coefficients.

** Q.4. How is binomial theorem used in real life?**: The Binomial theorem is used to establish results and solve problems in combinatorics, algebra, calculus and many other areas of mathematics. It is used to compare two large numbers, to find the remainder when a number to some large exponent is divided by a number and used in the probability to find the success or failure of an experiment. The binomial theorem is also used in weather forecasting, predicting the national economy in the next few years and distribution of IP addresses.

Ans

** Q.5. How is binomial theorem used in economics?**: Economists utilised the binomial theorem to count probabilities based on a large number of widely distributed factors in order to forecast how the economy will perform in the next years.

Ans

*We hope you find this article on Binomial Theorem* *helpful. In case of any queries, you can reach back to us in the comments section, and we will try to solve them. *