General Term in Binomial Expansion: When binomial expressions are raised to the power of \(2\) and \(3\) such as \((a + b)^2\) and \((p – q)^3\), we use a set of algebraic identities to find the expansion. The binomial expansions of these expressions are listed below:
1. \({\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\)
2. \({\left( {p – q} \right)^3} = {p^3} – {q^3} – 3{p^2}q + 3p{q^2}\)

What happens if the exponents of binomial expressions are more than \(3\)? Although we can calculate the expansion by repeated multiplication, it gets laborious. This is the reason we employ the binomial expansion formula. As the name suggests, when binomial expressions are raised to a power or degree, they have to be expanded and simplified by calculations. We use the binomial theorem to simplify this series of calculations.

Terms in an Expression

A term is a single mathematical expression. It may be positive or negative. A term may be a number, a variable or a combination of both. When a term has more than one variable, they are always multiplied, never added or subtracted. A term may have both a number and a variable. In such cases, the number is placed in front of the variable and is called it’s coefficient.
Examples of terms:
i. \(x,\,y,\,xy,\,{x^2}{y^2}\) are all single terms with variables \(x\), and \(y\).
ii. \(3p\) is also a term. Here, \(3\) is the coefficient, and the variable is \(p\).
iii. \(- 155 ab^2 c^3\) is also a term. Here, \(- 155\) is the coefficient, and the variables are \(ab^2 c^3\).

Binomial Expression

An algebraic expression that has two terms with a plus \((+)\) or minus \((-)\) sign is called a binomial expression. Examples of binomial expressions:
1. \(a + b \to \) the terms are \(a\) and \(b\)
2. \(2p – 3q \to \) the terms are \(2p\) and \(- 3q\)
3. \({x^2} – {y^2} \to \) the terms are \(x^2\) and \(- y^2\)
To add two or more binomials, the like terms are combined; and to multiply two binomials, we use the distributive property.

Pascal’s Triangle

Observe this table of binomial expansions.

Binomial Expression	Exponent	Expansion	Number of terms
\({\left( {a + b} \right)^1}\)	\(1\)	\({a + b}\)	\(2\)
\({\left( {a + b} \right)^2}\)	\(2\)	\({a^2} + 2ab + {b^2}\)	\(3\)
\({\left( {a + b} \right)^3}\)	\(3\)	\({a^3} + 3{a^2}b + 3a{b^2} + {b^3}\)	\(4\)
\({\left( {a + b} \right)^n}\)	\(n\)		\(n + 1\)

From this table, we know that there are \(n + 1\) terms in the expansion of \(( a + b)\) is raised to the power of \(n\) as \((a + b)^n\). Consider the coefficients of the terms in the expansion. They can be arranged to form a triangle as shown below.

Binomial Expansion	Coefficients

Observe that there is \(1\) at the two sides of the triangle. This triangle of coefficients is called Pascal’s Triangle.

Pascal’s Triangle and Binomial Expansion

Pascal’s triangles give us the coefficients of the binomial expansion of the form \((a + b)^n\) in the \({n^{{\rm{th}}}}\) row in the triangle. Each coefficient is achieved by adding two coefficients in the previous row, on the immediate left and immediate right. Now that we know to find the coefficients, how can we know the variables and their exponents, introduce – binomial theorem.

Binomial Theorem Example

For real numbers, \(a\) and \(b\), and a positive integer \(n\),
\({\left( {a + b} \right)^n}{ = ^n}{C_0}{a^n}{b^0}{ + ^n}{C_1}{a^{n – 1}}{b^1}{ + ^n}{C_2}{a^{n – 2}}{b^2}{ + ^n}{C_3}{a^{n – 3}}{b^3} + \ldots { + ^n}{C_n}{a^0}{b^n}\)
where \(C\) is called combination and is defined as:
\({}^n{C_r} = \frac{{n!}}{{r!\left( {n – r} \right)!}}\) for \(0 \le r \le n.\)
This is the expression that represents binomial expansion.

Features of Binomial Theorem

1. In an expansion of \((a + b)^n\), there are \((n + 1)\) terms.
2. \(n\) is a positive integer and is always greater than \(r\).
3. In any term in the expansion, the sum of powers of \(a\) and \(b\) is equal to \(n\).
(a) The exponent of \(b\) starts from \(0\) and increases with each term until it reaches the highest value of \(n\).
(b) The exponent of \(a\) starts from the highest value of \(n\), and decreases with each term until it reaches \(0\).
4. \({}^n{C_0},\,{}^n{C_1},\,{}^n{C_2},…{}^n{C_n}\) are called binomial coefficients.
5. The coefficients in the expansion are symmetric about the middle term; this means that the first coefficient is equal to the last. That is \(^n{C_0}{ = ^n}{C_n},{\,^n}{C_1}{ = ^n}{C_{n – 1}},{\,^n}{C_2}{ = ^n}{C_{n – 2}}\) and so on.

General Term in Binomial Expansion

Observe the terms in the binomial theorem for expansion.
\({\left( {a + b} \right)^n}{ = ^n}{C_0}{a^n}{b^0}{ + ^n}{C_1}{a^{n – 1}}{b^1}{ + ^n}{C_2}{a^{n – 2}}{b^2}{ + ^n}{C_3}{a^{n – 3}}{b^3} + \ldots { + ^n}{C_n}{a^0}{b^n}\)
Here,
Term \(1\) \( \to {T_1}{ = ^n}{C_0}{a^n}{b^0} = {a^n}\)
Term \(2\) \( \to {T_2}{ = ^n}{C_1}{a^{n – 1}}{b^1}\)
Term \(3\) \( \to {T_3}{ = ^n}{C_2}{a^{n – 2}}{b^2}\)
Term \(n\) \( \to {T_n}{ = ^n}{C_{n – 1}}{a^1}{b^{n – 1}}\)
Term \((n + 1)\) \( \to {T_{n + 1}}{ = ^n}{C_n}{a^0}{b^n} = {b^n}\)
This can be generalized to get the formula for the \((r + 1)^{th}\) or the general term.

Formula for General Term in Binomial Expansion

In a binomial expansion to the power \(n,\) there are \((n + 1)\) terms. The general term or the \((r + 1)^{th}\) term in a binomial expansion is given by:
\({T_{r + 1}}{ = ^n}{C_r}{a^{n – r}}{b^r}\)
where, \(r \le 0 \le n.\)

Middle Term in Binomial Expansion

In the binomial expansion of \((a + b)^n\), there are \(n + 1\) terms. The number of the middle term will vary based on whether \(n\) is even or odd.

i. For even values of n
If \(n\) is an even number, then the expansion will have an odd number of terms. Then the middle term will be the \({\left( {\frac{n}{2} + 1} \right)^{th}}\) term.
For example, on expanding \((a + b)^2\) we get \(3\) terms. Hence, the middle term will be:
\({\left( {\frac{n}{2} + 1} \right)^{th}} = \left( {\frac{2}{2} + 1} \right) = \left( {1 + 1} \right) = {2^{nd}}\) term.

ii. For odd values of n
If \(n\) is an odd number, then the expansion will have an even number of terms. Hence, the expansion will have two middle terms. The middle term will be the \({\left( {\frac{{n + 1}}{2}} \right)^{th}}\) term and the \({\left( {\frac{{n + 3}}{2}} \right)^{th}}\) term.
For example, on expanding \((a + b)^7\) we get \(8\) terms. Hence, the two middle terms will be:
\({\left( {\frac{{n + 1}}{2}} \right)^{th}} = \left( {\frac{{7 + 1}}{2}} \right) = \left( {\frac{8}{2}} \right) = {4^{th}}\) term
\({\left( {\frac{{n + 3}}{2}} \right)^{th}} = \left( {\frac{{7 + 3}}{2}} \right) = \left( {\frac{{10}}{2}} \right) = {5^{th}}\) term

Formula for Middle Term in Binomial Expansion

i. When n is even
Middle term \(= {\left( {\frac{n}{2} + 1} \right)^{th}}\) term
ii. When n is odd
There are two middle terms.
Middle term \(1 = {\left( {\frac{{n + 1}}{2}} \right)^{th}}\) term
Middle term \( 2= {\left( {\frac{{n + 3}}{2}} \right)^{th}}\) term

Signs of Binomial Expansion

For a binomial expansion, the coefficients can be derived using Pascal’s Triangle, while the variables and their exponents can be calculated using the binomial theorem.
The binomial theorem states that, for a real number \(a\) and \(b\), and a positive integer \(n\),
\({\left( {a + b} \right)^n}{ = ^n}{C_0}{a^n}{b^0}{ + ^n}{C_1}{a^{n – 1}}{b^1}{ + ^n}{C_2}{a^{n – 2}}{b^2}{ + ^n}{C_3}{a^{n – 3}}{b^3} + \ldots { + ^n}{C_n}{a^0}{b^n}\)
where \(C\) is called combination and is defined as: \(^n{C_r} = \frac{{n!}}{{r!\left( {n – r} \right)!}}\) for \(0 \le r \le n.\)
In the case of a subtraction binomial \(( a – b)^n\), while the first term of the expansion will be positive, the following terms will be alternatively negative. It can be represented as:
\({\left( {a – b} \right)^n}{ = ^n}{C_0}{a^n}{b^0}{ – ^n}{C_1}{a^{n – 1}}{b^1}{ + ^n}{C_2}{a^{n – 2}}{b^2}{ – ^n}{C_3}{a^{n – 3}}{b^3} + { \ldots ^n}{C_n}{a^0}{b^n}\).

Solved Examples: General Term in Binomial Expansion

Q.1. Find the general term of \((1 + x)^{10}\).
Ans: General Term, \({T_{r + 1}}{ = ^n}{C_r}{a^{n – r}}{b^r}\), \({{r}} = 0,\;1,\;2,\; \ldots \) \(^n{C_r} = \frac{{n!}}{{r!\left( {n – r} \right)!}}\)
Here,
\( a = 1\)
\(b = x\)
\( n = 10\)
Therefore, general term, \({T_{r + 1}}{ = ^{10}}{C_r}{\left( 1 \right)^{10 – r}}{x^r}{ = ^{10}}{C_r}{x^r}\), where \(0 \le r \le 10.\)

Q.2. For the expansion of \({\left( {x – 3{z^2}} \right)^5}\), find the general term and the coefficient of \({x^3}{z^4}\).
Ans: General Term, \({T_{r + 1}}{ = ^n}{C_r}{a^{n – r}}{b^r}\) \({{r}} = 0,\;1,\;2,\; \ldots \) \(^n{C_r} = \frac{{n!}}{{r!\left( {n – r} \right)!}}\)
Here,
\( a = x\)
\(b = – 3z^2\)
\(n = 5\)
Therefore, General Term, \({T_{r + 1}}{ = ^5}{C_r}{x^{5 – r}}{\left( { – 3{z^2}} \right)^r}\), where \(0 \le r \le 5.\)
Given variable: \({x^3}{z^4}\)
Equating the variables,
\({x^3}{z^4} = {x^{5 – r}}{\left( {{z^2}} \right)^r}\)
Comparing the powers of \(x\) or \(z\), we get \(r = 2\).
The term with \(r = 2,\,{T_{2 + 1}}{ = ^5}{C_2}{x^{5 – 2}}{\left( { – 3{z^2}} \right)^2}\)
\( = \frac{{5!}}{{2!\left( {5 – 2} \right)!}}{x^3}{\left( { – 3{z^2}} \right)^2}\)
\( = \frac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times \left( 3 \right)!}}{x^3}{\left( { – 3} \right)^2}{z^4}\)
\( = \frac{{5 \times 4 \times 3}}{{3 \times 2 \times 1}}{x^3} \times 9{z^4}\)
\( = 90{x^3}{z^4}\)
Coefficient of the \({x^3}{z^4} = 90\).

Q.3. Find the middle term of \({\left( {2x + 3y} \right)^6}\).
Ans: Here, \(n = 6\),
Hence, the expansion will have \( 6 + 1 = 7\) terms. \(7\) is an odd number.
Therefore, Middle term \({\left( {\frac{6}{2} + 1} \right)^{th}} = 3 + 1 = {4^{th}}\) term.
General Term, \({T_{r + 1}}{ = ^n}{C_r}{a^{n – r}}{b^r}\) \({{r}} = 0,\;1,\;2,\; \ldots \) \(^n{C_r} = \frac{{n!}}{{r!\left( {n – r} \right)!}}\)
Therefore, \({4^{th}}\) term \({T_{3 + 1}}{ = ^6}{C_3}{\left( {2x} \right)^{6 – 3}}{\left( {3y} \right)^3}\)
\( = \frac{{6!}}{{3!\left( {6 – 3} \right)!}}{\left( {2x} \right)^3}{\left( {3y} \right)^3}\)
\( = \frac{{6!}}{{3! \times 3!}}8{x^3} \times 27{y^3}\)
\( = \frac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{1 \times 2 \times 3 \times 1 \times 2 \times 3}}8 \times 27{x^3}{y^3}\)
\( = \frac{{6 \times 5 \times 4}}{{1 \times 2 \times 3}}216{y^3}\)
\( = 20 \times 216{x^3}{y^3}\)
Middle term, \({4^{th}}\) term \( = 4,320{x^3}{y^3}\).

Q.4. Find the middle term in the expansion of \({\left( {{x^4} – \frac{1}{{{x^3}}}} \right)^{11}}\)
Ans: Given: \({\left( {{x^4} – \frac{1}{{{x^3}}}} \right)^{11}} = {\left( {{x^4} + \left( { – \frac{1}{{{x^3}}}} \right)} \right)^{11}}\)
Here, \(n = 11\),
Hence, the expansion will have \(11 + 1 = 12\) terms. \(12\) is an even number.
Therefore, there are \(2\) middle terms.
Middle term 1 \( = {\left( {\frac{{11 + 1}}{2}} \right)^{th}} = \frac{{12}}{2} = {6^{th}}\) term
Middle term 2 \( = {\left( {\frac{{11 + 3}}{2}} \right)^{th}} = \frac{{14}}{2} = {7^{th}}\) term
General Term, \({T_{r + 1}}{ = ^n}{C_r}{a^{n – r}}{b^r}\) \({{r}} = 0,\;1,\;2,\; \ldots \) \(^n{C_r} = \frac{{n!}}{{r!\left( {n – r} \right)!}}\)
Here,
\(a = x^4\)
\(b = – \frac {1}{x^2}\)

6th term	7th term
6th term, \({T_{5 + 1}}{ = ^{11}}{C_5}{({x^4})^{11 – 5}}{\left( {\frac{1}{{{x^3}}}} \right)^5}\) \( = \frac{{11!}}{{5!\left( {11 – 5} \right)!}}{x^{4 \times 6}}{\left( {\frac{1}{{{x^3}}}} \right)^5}\) \( = \frac{{11!}}{{5!6!}}{x^{24}} \times \frac{1}{{{x^{15}}}}\) \( = 462 \times {x^{24 – 15}}\) \( = 462{x^9}\)	7th term \({T_{6 + 1}}{ = ^{11}}{C_6}{({x^4})^{11 – 6}}{\left( {\frac{1}{{{x^3}}}} \right)^6}\) \( = \frac{{11!}}{{6!\left( {11 – 6} \right)!}}{x^{4 \times 5}}{\left( {\frac{1}{{{x^3}}}} \right)^6}\) \( = \frac{{11!}}{{6!5!}}{x^{20}} \times \frac{1}{{{x^{18}}}}\) \( = 462 \times {x^{20 – 18}}\) \( = 462{x^2}\)

Hence, the middle terms of the expansion are \( = 462{x^9}\) and \( = 462{x^2}\).

Q.5. If the coefficient of the third term in the expansion of \({\left( {x – \frac{1}{4}} \right)^n}\) is \(\frac{{33}}{8}\). Find the middle term.
Ans: Given: \({\left( {x – \frac{1}{4}} \right)^n} = {\left( {x + \left( { – \frac{1}{4}} \right)} \right)^n}\)
General Term, \({T_{r + 1}}{ = ^n}{C_r}{a^{n – r}}{b^r}\), \({{r}} = 0,\;1,\;2,\; \ldots \); \(^n{C_r} = \frac{{n!}}{{r!\left( {n – r} \right)!}}\)
Third term, \({T_{2 + 1}}{ = ^3}{C_2}{a^{n – 2}}{b^2} = \frac{{33}}{8}\)
Here,
\(a = x\)
\(b = – \frac {1}{4}\)
Equating the coefficients,
\(\therefore \,\frac{{n!}}{{2!\left( {n – 2} \right)!}}{\left( { – \frac{1}{4}} \right)^2} = \frac{{33}}{8}\)
\(\frac{{n!}}{{2 \times \left( {n – 2} \right)!}} \times \frac{1}{{16}} = \frac{{33}}{8}\)
\(\frac{{n!}}{{2 \times \left( {n – 2} \right)!}} = \frac{{33}}{8} \times 16\)
\(\frac{{n\left( {n – 1} \right)\left( {n – 2} \right)!}}{{\left( {n – 2} \right)!}} = 66 \times 2\)
\(n\left( {n – 1} \right) = 132\)
\({n^2} – n – 132 = 0\)
Solving for \(n\),
\(\left( {n + 11} \right)\left( {n – 12} \right) = 0\)
\(\therefore \,n = – 11\) or \(12\)
Since, \(n\) is a positive integer, \(= 12\).
Here, \(n\) is even, hence there is only one middle term.
Middle term \({\left( {\frac{n}{2} + 1} \right)^{th}}\) term \({\left( {\frac{{12}}{2} + 1} \right)^{th}} = 6 + 1 = {7^{th}}\) term
General Term, \({T_{r + 1}}{ = ^n}{C_r}{a^{n – r}}{b^r}\) \({{r}} = 0,\;1,\;2,\; \ldots \) \(^n{C_r} = \frac{{n!}}{{r!\left( {n – r} \right)!}}\)
\({7^{th}}\) term, \({T_{6 + 1}}{ = ^{12}}{C_6}{x^{12 – 6}}{\left( { – \frac{1}{4}} \right)^6}\)
\( = \frac{{12!}}{{6!\left( {12 – 6} \right)!}}{x^{12 – 6}}{\left( { – \frac{1}{4}} \right)^6}\)
\( = \frac{{12!}}{{6! \times 6!}}{x^{12 – 6}}{\left( { – \frac{1}{4}} \right)^6}\)
\( = \frac{{12!}}{{6! \times 6!}}{\left( {\frac{1}{4}} \right)^6}{x^6}\)
\( = 924 \times \frac{1}{{4096}}{x^6}\)
\({7^{th}}\) term, \( = \frac{{231}}{{1024}}{x^6}\) is the middle term.

Summary

This article helps understand the general term in binomial expansion by explaining terms in an expression, followed by Pascal’s triangle to help identify the coefficients in binomial expansion. The variables in the expansion can be achieved using the Binomial Theorem. It also lists and explains the formula for the general term in a binomial expansion. Lastly, it explains how to decide the signs in an expansion. There are also a few solved examples to help gain clarity on the topic and calculations involved.

Learn Concepts of Binomial Distribution

Frequently Asked Questions (FAQs)

Q.1. What are terms in binomials?
Ans: A term is a single mathematical expression. It may be positive or negative. A term may be a number, a variable or a combination of both. When a term has more than one variable, they are always multiplied, never added or subtracted. A term may have both a number and a variable. For example, \(a,\;b^2,\) and \(-3cp\).

Q.2. What is the general term in a binomial expansion?
Ans: The general term or the \((r + 1)^{th}\) term in a binomial expansion is given by:
\({T_{r + 1}}{ = ^n}{C_r}{a^{n – r}}{b^r},\)\(\forall\, r \le 0 \le n.\)

Q.3. How do you find a term in a binomial expansion?
Ans: Step 1: Identify \(n\). It is the power of the binomial to be expanded.
Step 2: Identify the number of the term to be calculated. This is \((r + 1)\).
Step 3: Calculate \(r\).
Step 4: identify \(a\) and \(b\) from the binomial.
Step 5: Substitute and simply the formula \({T_{r + 1}}{ = ^n}{C_r}{a^{n – r}}{b^r}\).

Q.4. How many terms are in a binomial expansion?
Ans: In a binomial expansion of \((a + b)^n\), there are \((n + 1)\) terms.

Q.5. How do you find the middle term in a binomial expansion?
Ans: Step 1: Determine the middle term in an expansion.
i. When n is even
Middle term \( = {\left( {\frac{n}{2} + 1} \right)^{th}}\) term
ii. When n is odd
There are two middle terms.
Middle term 1 \(= {\left( {\frac{{n + 1}}{2}} \right)^{th}}\) term
Middle term 2 \(= {\left( {\frac{{n + 3}}{2}} \right)^{th}}\) term
Step 2: Use the number of the middle term obtained from Step 1 in the formula:
\({T_{r + 1}} = {}^n{C_r}{a^{n – r}}{b^r};\,r = 0,\,1,\,2,\,…;\,{}^n{C_r} = \frac{{n!}}{{r!\left( {n – r} \right)!}}.\)

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