Binomial Theorem for Positive Integral Indices: The algebraic expression of the form \(a+b\) is called a binomial expression. Although in principle it is easy to raise \(a+b\) to any power, raising it to a very high power could be tedious. This will then involve a tedious process of performing repeated multiplication of the expressions. But such high powers of binomial expression are used to find roots of equations in higher powers.

This article explains how to derive the expansion of \({(a + b)^n}\), for any natural number \(n\) using the binomial theorem. It is a quick method to expand binomial expressions to huge powers without such long calculations.

Learn about Binomial Distribution here

Decoding Binomial Expansion

To find a pattern in the expansion of  \({(a + b)^n}\), first look at the following expansions involving small positive integral indices.

\({(a + b)^1} = a + b\)
\({(a + b)^2} = {a^2} + 2ab + {b^2}\)
\({(a + b)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}\)
\({(a + b)^4} = {(a + b)^3}(a + b) = {a^4} + 4{a^3}b + 6{a^2}{b^2} + 4a{b^3} + {b^4}\)

 In all the above expansions, observe that:

• The total number of terms in the expansion is always \(1\) more than the index.
  • For example, in the expansion of \({(a + b)^2}\), number of terms is \(3\) whereas the index of \({(a + b)^2}\) is \(2\).
• Powers of the first term \({\rm{‘}}{a^\prime }\) go on decreasing by \(1\), whereas the powers of the second term \({\rm{‘}}{b^\prime }\) increase by \(1\) in the successive terms.
• In each term of the expansion, the sum of the indices of \(a\) and \(b\) is the same as index of \((a + b)\).

Extrapolating, notice how the exponents of \(a\) and \(b\) behave in the expansion of \({(a + b)^5}\).

Pascal’s Triangle

Now, arrange the coefficients in the above binomial expansions in a triangle as shown below.

This triangular arrangement of coefficients with \(1s\) at the top vertex and along the two slanting sides is known as Pascal’s triangle. This is named after a French mathematician, Blaise Pascal. It is referred to as Meru Prastara by Pingla in Indian manuscripts.

Statement of Binomial Theorem for Positive Integral Indices

The Binomial Theorem for Positive Integral Indices states that, “The total number of terms in the expansion of  \({(a + b)^n}\) is one more than the index, \(n\). So, in the expansion of  \({(a + b)^n}\), the number of terms is \(n+1\), where \(n\) is the index, and is a positive integer.

Properties of Binomial Theorem

The properties of Binomial Theorem:

• There are \(n+1\) terms in the expansion, the first being \({a^n}\) and the last being \({b^n}\).
• The exponents of \(a\) decrease by \(1\) from \(0\) term to term, while the exponents of \(b\) increase by \(1\).
• The sum of the exponents of \(a\) and \(b\) in each term is \(n\).
• Coefficients of the terms in binomial expansion equidistant from both the ends are equal. i.e., coefficients are symmetric.

Key Property: Every entry (other than \(1\)) is the sum of the two entries diagonally above it. For instance, we find the sixth and seventh rows, starting with the fifth row:

Limitations of Pascal’s Triangle

Pascal’s triangle is useful in finding the binomial expansions for reasonably small values of \(n\), it isn’t practical for finding expansions for large values of \(n\).

For example, to find the \({100^{th}}\) row of this triangle, one must also find the entries of the first \(99\) rows.

To overcome this limitation, a formula is used that allows the calculation of any coefficient in the binomial expansion without using Pascal’s Theorem.

Binomial Coefficients

Let \(n\) and \(r\) be non-negative integers, The binomial coefficient is denoted by \({n_{{C_r}}}\).

\({n_{{C_r}}} = \frac{{n!}}{{(n – r)!r!}}\), where \(0 \le r \le n\)

The Pascal’s triangle can now be rewritten as,

Therefore, for the index \(8\) the row would be:

\(\begin{array}{*{20}{l}}
{{8_{{C_0}}}}&{{8_{{C_1}}}}&{{8_{{C_2}}}}&{{8_{{C_3}}}}&{{8_{{C_4}}}}&{{8_{{C_5}}}}&{{8_{{C_6}}}}&{{8_{{C_7}}}}&{{8_{{C_8}}}}
\end{array}\)

Properties of Binomial Coefficients

1. Consider

\(^n{C_{n – r}} = \frac{{n!}}{{(n – r)![n – (n – r)]!}}\)
\( = \frac{{n!}}{{(n – r)!r!}}\)
\({ = ^n}{C_r}\)

Thus, \(^n{C_{n – r}}{ = ^n}{C_r}\), for \(0 \le r \le n\)

2. \(^n{C_0} = \frac{{n!}}{{0!(n – 0)!}}\)
\( = \frac{{n!}}{{n!}},\quad 0! = 1\)
\(=1\)
\({\therefore ^n}{C_0} = 1,\forall n\)

3. If \(^n{C_r}{ = ^n}{C_s}\), then either \(s=r\) or \(s=n-r\)
Proof:
Let \(r \equiv s\) & \(r>s\),
Then, \(n – r < n – s\)
As \(\frac{{n!}}{{(n – r)!r!}} = \frac{{n!}}{{(n – s)!s!}} \Rightarrow (n – s)!s! = (n – r)!r!\)
\( \Rightarrow s!(s + 1)(s + 2) \ldots \ldots r(n – r)!\)
\( = s!(n – r)!(n – r + 1)(n – r + 2) \ldots (n – s) \Rightarrow (s + 1)(s + 2) \ldots \ldots r\)
\( = (n – r + 1)(n – r + 2) \ldots \ldots (n – s)\)

4. \(^n{C_r}{ + ^n}{C_{r – 1}}{ = ^{n + 1}}{C_r}\)
Proof:
L.H.S: \(^n{C_r}{ + ^n}{C_{r – 1}} = \frac{{n!}}{{(n – r)!r!}} + \frac{{n!}}{{(n – r + 1)!(r – 1)!}}\)
\( = n!\left[ {\frac{1}{{(n – r)!r!}} + \frac{1}{{(n – r + 1)!(r – 1)!}}} \right]\)
\( = n!\left[ {\frac{1}{{(n – r)!r(r – 1)!}} + \frac{1}{{(n – r + 1)!(r – 1)!}}} \right]\)
\( = \frac{{n!}}{{(r – 1)!}}\left[ {\frac{1}{{(n – r)!r}} + \frac{1}{{(n – r + 1)!}}} \right]\)
\( = \frac{{n!}}{{(r – 1)!}}\left[ {\frac{1}{{r(n – r)!}} + \frac{1}{{(n – r + 1)(n – r)!}}} \right]\)
\( = \frac{{n!}}{{(r – 1)!(n – r)!}}\left[ {\frac{1}{r} + \frac{1}{{(n – r + 1)}}} \right] = \frac{{n!}}{{(r – 1)!(n – r)!}}\left[ {\frac{{n – r + 1 + r}}{{r(n – r + 1)}}} \right]\)
\( = \frac{{n!}}{{r(r – 1)!(n – r)!}}\left[ {\frac{{n + 1}}{{n – r + 1}}} \right] = \frac{{(n + 1)n!}}{{r!(n – r + 1)(n – r)!}}\)
\( = \frac{{(n + 1)!}}{{(n + 1 – r)!r!}}{ = ^{n + 1}}{C_r}\):R.H.S.

5. \(^n{C_0}{ + ^n}{C_1} + { \cdots ^n}{C_n} = {2^n}\)
We know that, \({(1 + x)^n}{ = ^n}{C_0}{ + ^n}{C_1}x{ + ^n}{C_2}{x^2} + \cdots \ldots { + ^n}{C_n}{x^n}\)
If \(x = 1,{(1 + 1)^n}{ = ^n}{C_0}{ + ^n}{C_1} \times 1{ + ^n}{C_2} \times {1^2} + \cdots ..{ + ^n}{C_n} \times {1^n}\)
\({2^n} = 1{ + ^n}{C_1}{ + ^n}{C_2} + \cdots \ldots { + ^n}{C_n}\)

6. \(^n{C_0}{ + ^n}{C_2}{ + ^n}{C_4} + \cdots { = ^n}{C_1}{ + ^n}{C_3}{ + ^n}{C_5} + \cdots = {2^{(n – 1)}}\)
Proof:
We know that, \({(1 + x)^n}{ = ^n}{C_0}{ + ^n}{C_1}x{ + ^n}{C_2}{x^2} + \cdots ..{ + ^n}{C_n}{x^n}\)
If, \(x = 1,{(1 + 1)^n}{ = ^n}{C_0}{ + ^n}{C_1} \times 1{ + ^n}{C_2} \times {1^2} + \cdots \ldots { + ^n}{C_n} \times {1^n}\)
\({2^n} = 1{ + ^n}{C_1}{ + ^n}{C_2} + \cdots \ldots { + ^n}{C_n}\)
\(^n{C_1}{ + ^n}{C_2} + \cdots \ldots { + ^n}{C_n} = {2^n} – 1\)

7. \(^n{C_r} = {\left( {\frac{n}{r}} \right)^{n – 1}}{C_{r – 1}} = \left( {\frac{n}{r}} \right){\left( {\frac{{n – 1}}{{r – 1}}} \right)^{n – 2}}{C_{r – 2}} = \cdots \)

8. \(^n{C_r}\) has maximum value if
a. \(r = \frac{n}{2}\) when \(n\)is even
b. \(r = \frac{{n – 1}}{2}\) or \(\frac{{n + 1}}{2}\) when \(n\) is odd.

Binomial Theorem for any Positive Integral Index \(n\)
Theorem: If \(a,b \in R\), and \(n \in N\), then
\({(a + b)^n}{ = ^n}{C_0}{a^n}{ + ^n}{C_1}{a^{n – 1}}b{ + ^n}{C_2}{a^{n – 2}}{b^2} + \cdots { + ^n}{C_{n – 1}}a \cdot {b^{n – 1}}{ + ^n}{C_n}{b^n}\)
Proof:
\(P(n):{(a + b)^n}{ = ^n}{C_0}{a^n}{ + ^n}{C_1}{a^{n – 1}}b{ + ^n}{C_2}{a^{n – 2}}{b^2} + \cdots { + ^n}{C_{n – 1}}a \cdot {b^{n – 1}}{ + ^n}{C_n}{b^n}\)
For \(n = 1,P(1):{(a + b)^1}{ = ^1}{C_0}{a^1}{ + ^1}{C_1}{b^1} = a + b\)
Hence, \(P(1)\) is true.
Suppose \(P(k)\) is true for some positive integer \(k\) then:
\({(a + b)^k}{ = ^k}{C_0}{a^k}{ + ^k}{C_1}{a^{k – 1}}b{ + ^k}{C_2}{a^{k – 2}}{b^2} + \cdots { + ^k}{C_k}{b^k} \ldots (1)\)
We shall prove that \(P(k + 1)\) is also true, i.e.
\({(a + b)^{k + 1}}{ = ^{k + 1}}{C_0}{a^{k + 1}}{ + ^{k + 1}}{C_1}{a^k}b{ + ^{k + 1}}{C_2}{a^{k – 1}}{b^2} + \cdots { + ^{k + 1}}{C_{k + 1}}{b^{k + 1}}\)
Now, \({(a + b)^{k + 1}} = (a + b){(a + b)^k}\)
\( = (a + b)\left( {^k{C_0}{a^k}{ + ^k}{C_1}{a^{k – 1}}b{ + ^k}{C_2}{a^{k – 2}}{b^2} + \cdots { + ^k}{C_{k – 1}}a{b^{k – 1}}{ + ^k}{C_k}{b^k}} \right)\)
\({ = ^k}{C_0}{a^{k + 1}}{ + ^k}{C_1}{d^k}b{ + ^k}{C_2}{a^{k – 1}}{b^2} + \cdots { + ^k}{C_{k – 1}}{a^2}{b^{k – 1}}{ + ^k}{C_k}a{b^k}{ + ^k}{C_0}{a^k}b{ + ^k}{C_1}{a^{k – 1}}{b^2}{ + ^k}{C_2}{a^{k – 2}}{b^3} + \cdots { + ^k}{C_{k – 1}}a{b^k}{ + ^k}{C_k}{b^{k + 1}}\)
\({ = ^k}{C_0}{a^{k + 1}} + \left( {^k{C_1}{ + ^k}{C_0}} \right){a^k}b + \left( {^k{C_2}{ + ^k}{C_1}} \right){a^{k – 1}}{b^2} + \cdots + \left( {^k{C_k}{ + ^k}{C_{k – 1}}} \right)a{b^k}{ + ^k}{C_k}{b^{k + 1}}\)
by using, \(^{k + 1}{C_0} = 1{,^k}{C_r}{ + ^k}{C_{r – 1}}{ = ^{k + 1}}{C_r}\) and \(^k{C_k} = 1{ = ^{k + 1}}{C_{k + 1}}{ = ^{k + 1}}{C_0}{a^{k + 1}}{ + ^{k + 1}}{C_1}{a^k}b{ + ^{k + 1}}{C_2}{a^{k – 1}}{b^2} + \cdots { + ^{k + 1}}{C_k}a{b^k}{ + ^{k + 1}}{C_{k + 1}}{b^{k + 1}}\)
Hence, it is proved that \(P(k+1)\) is true whenever \(P(k)\) is true.
Therefore, by principle of mathematical induction, \(P(n)\) is true for every positive integer \(n\).

Important Observations:

1. The notation \(\sum_{k=0}^{n}{ }^{n} C_{k} a^{n-k} b^{k}\) stands for

\({ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}\)

So, the theorem is stated as \((a+b)^{n}=\sum_{k=0}^{n}{ }^{n} C_{k} a^{n-k} b^{k}\).

2. The coefficients \({ }^{n} C_{r}\) occurring in the binomial theorem are known as binomial coefficients.

3. There are \((n+1)\) terms in the expansion of \((a+b)^{n}\), i.e., one more than the index.

4. In the successive terms of the expansion the index of \(a\) goes on decreasing by unity. It is \(n\) in the first term, \((n-1)\) in the second term, and so on ending with zero in the last term. At the same time the index of \(b\) increases by unity, starting with zero in the first term, \(1\) in the second and so on ending with \(n\) in the last term.

5. In the expansion of \((a+b)^{n}\), the sum of the indices of \(a\) and \(b\) is \(n+0=n\) in the first term, \((n-1)+1=n\) in the second term and so on \(0+n=n\) in the last term. Thus, it can be seen that the sum of the indices of \(a\) and \(b\) is \(n\) in every term of the expansion.

Solved Examples

Q.1. Find the value of (i) \({ }^{7} C_{3}\) (ii) \({ }^{10} C_{7}\) (iii) \({ }^{52} C_{3}\)
Ans: We know that, \(n C_{r}=\frac{n !}{r !(n-r) !}\)

(i) \({ }^{7} C_{3}=\frac{7 !}{3 !(7-3) !}\)

\(=\frac{7 !}{3 ! 4 !}\)

\(=\frac{7 \times 6 \times 5 \times 4 !}{3 \times 2 \times 1 \times 4 !}\)

\(=\frac{7 \times 6 \times 5}{3 \times 2 \times 1}\)

\({ }^{7} C_{3}=35\)

(ii) \({ }^{10} C_{7}=\frac{10 !}{7 !(10-7) !}\)

\(=\frac{10 !}{7 !(10-7) !}\)

\(=\frac{10 \times 9 \times 8}{3 \times 2}\)

\({ }^{10} C_{7}=120\)

(iii) \({ }^{52} C_{3}=\frac{52 !}{3 !(52-3) !}\)

\(=\frac{52 !}{3 !(49) !}\)

\(=\frac{52 \times 51 \times 50}{3 \times 2 \times 1}\)

\({ }^{52} C_{3}=22100\)

Q.2. Use the Binomial Theorem to expand \(\left(x^{2}+3 y\right)^{5}\).
Ans: Here,

\(a=x^{2}\)

\(b=3 y\)

\(n=5\)

Using binomial theorem for positive integral index, we have,

\({\left( {{x^2} + 3y} \right)^5}{ = ^5}{C_0}{\left( {{x^2}} \right)^5}{(3y)^0}{ + ^5}{C_1}{\left( {{x^2}} \right)^4}{(3y)^1}{ + ^5}{C_2}{\left( {{x^2}} \right)^3}{(3y)^2}{ + ^5}{C_3}{\left( {{x^2}} \right)^2}{(3y)^3}{ + ^5}{C_4}{\left( {{x^2}} \right)^1}{(3y)^4}{ + ^5}{C_5}{\left( {{x^2}} \right)^0}{(3y)^5}\)

Here,

\({ }^{5} C_{0}={ }^{5} C_{5}=1\)

\({ }^{5} C_{1}={ }^{5} C_{4}=5\)

\({ }^{5} C_{2}={ }^{5} C_{3}=\frac{5 \times 4}{2 \times 1}=10\)

\(\therefore {\left( {{x^2} + 3y} \right)^5} = 1\left( {{x^{10}}} \right)(1) + 5\left( {{x^8}} \right)(3y) + 10\left( {{x^6}} \right)\left( {9{y^2}} \right) + 10\left( {{x^4}} \right)\left( {27{y^3}} \right) + 5\left( {{x^2}} \right)\left( {81{y^4}} \right) + 1(1)\left( {243{y^5}} \right)\)

\(\therefore\left(x^{2}+3 y\right)^{5}=x^{10}+15 x^{8} y+90 x^{6} y^{2}+270 x^{4} y^{3}+405 x^{2} y^{4}+243 y^{5}\)

Q.3. Expand \(\left(x^{2}+\frac{3}{x}\right)^{4}, x \neq 0\) using the binomial theorem.
Ans: Using binomial theorem for positive integral index, we have,

\(\left(x^{2}+\frac{3}{x}\right)^{4}={ }^{4} C_{0}\left(x^{2}\right)^{4}\left(\frac{3}{x}\right)^{0}+{ }^{4} C_{1}\left(x^{2}\right)^{3}\left(\frac{3}{x}\right)^{1}{ }^{4} C_{2}\left(x^{2}\right)^{2}\left(\frac{3}{x}\right)^{2}+{ }^{4} C_{3}\left(x^{2}\right)\left(\frac{3}{x}\right)^{3}+{ }^{4} C_{4}\left(\frac{3}{x}\right)^{4}\)

\(=x^{8}+4 x^{6} \cdot \frac{3}{x}+6 x^{4} \cdot \frac{9}{x^{2}}+4 x^{2} \cdot \frac{27}{x^{3}}+\frac{81}{x^{4}}\)

\(=x^{8}+12 x^{5}+54 x^{2}+\frac{108}{x}+\frac{81}{x^{4}}\)

Q.4. Use the Binomial Theorem to expand \((\sqrt{5}+\sqrt{3})^{4}\).
Ans: Here,

\(a=\sqrt{5}\)

\(b=\sqrt{3}\)

\(n=4\)

Using binomial theorem for positive integral index, we have,

\({(\sqrt 5 + \sqrt 3 )^4}{ = ^4}{C_0}{(\sqrt 5 )^4}{(\sqrt 3 )^0}{ + ^4}{C_1}{(\sqrt 5 )^3}{(\sqrt 3 )^1}{ + ^4}{C_2}{(\sqrt 5 )^2}{(\sqrt 3 )^2}{ + ^4}{C_3}{(\sqrt 5 )^1}{(\sqrt 3 )^3}{ + ^4}{C_4}{(\sqrt 5 )^0}{(\sqrt 3 )^4}\)

Here,

\({ }^{4} C_{0}={ }^{4} C_{4}=1\)

\({ }^{4} C_{1}={ }^{4} C_{3}=4\)

\({ }^{4} C_{2}=\frac{4.3}{2.1}=6\)

\(\Rightarrow(\sqrt{5}+\sqrt{3})^{4}=1(25)(1)+4(5 \sqrt{5})(3 \sqrt{3})+6(5)(3)+4(\sqrt{5})(3 \sqrt{3})+1(1)(9)\)

\(=25+20 \sqrt{15}+90+12 \sqrt{15}+9\)

\(\therefore(\sqrt{5}+\sqrt{3})^{4}=124+32 \sqrt{15}\)

Q.5. Which is larger \((1.01)^{1000000}\) or \(10,000\) ?
Ans: Splitting \(1.01\) and using binomial theorem to write the first few terms we have

\((1.01)^{1000000}=(1+0.01)^{1000000}\)

\(={ }^{1000000} C_{0}+{ }^{1000000} C_{1}(0.01)+\) other positive terms

\(=1+1000000 \times 0.01+\) other positive terms

\(=1+10000+\) other positive terms

\(>10000\)

Hence \((1.01)^{1000000}>10000\)

Q.6. Using binomial theorem, prove that \(6^{n}-5n\) always leaves remainder \(1\) when divided by \(25\).
Ans: For two numbers \(a\) and \(b\) if we can find numbers \(q\) and \(r\) such that \(a=b q+r\), then we say that \(b\) divides \(a\) with \(q\) as quotient and \(r\) as remainder. Thus, in order to show that \(6^{n}-5 n\) leaves remainder \(1\) when divided by \(25\), we prove that

\(6^{n}-5 n=25 k+1\), where \(k\) is some natural number.

We have

\((1+a)^{n}={ }^{n} C_{0}+{ }^{n} C_{1} a+{ }^{n} C_{2} a^{2}+\ldots+{ }^{n} C_{n} a^{n}\)

For \(a=5\), we get

\((1+5)^{n}={ }^{n} C_{0}+{ }^{n} C_{1} 5+{ }^{n} C_{2} 5^{2}+\ldots+{ }^{n} C_{n} 5^{n}\)

\((6)^{n}=1+5 n+5^{2} \cdot{ }^{n} C_{2}+5^{3} \cdot{ }^{n} C_{3}+\ldots+5^{n}\)

\(6^{n}-5 n=1+5^{2}\left({ }^{n} C_{2}+{ }^{n} C_{3} 5+\ldots+5^{n-2}\right)\)

or \(6^{n}-5 n=1+25\left({ }^{n} C_{2}+5 \cdot{ }^{n} C_{3}+\ldots+5^{n-2}\right)\)

or \(6^{n}-5 n=25 k+1\) where \(k={ }^{n} C_{2}+5 \cdot{ }^{n} C_{3}+\ldots+5^{n-2}\).

This shows that when \(6^{n}-5 n\) is divided by \(25\), leaves a remainder of \(1\).

Summary

The algebraic expression of the form \(a+b\) is called a binomial expression. The formula that gives the expansion of \((a+b)^{n}\) for any natural number \(n\) is called the Binomial Theorem for positive integral indices. Using this theorem helps us avoid tedious multiplications in order to expand huge powers of \((a+b)^{n}\). Also discussed are Pascal’s triangle and its properties along with their proofs. The article also explains binomial coefficients, their properties and how to calculate coefficients in specific binomial expansions using the formula \(n_{C_{r}}=\frac{n !}{(n-r) ! r !}\), where \(0 \leq r \leq n\).

Frequently Asked Questions (FAQs) – Binomial Theorem for Positive Integral Indices

Q.1. What do you mean by Binomial Theorem?
Ans: An expression consisting of two different terms is called a binomial expression. For all values of \(a\) and \(b\), the algebraic expression of the form \(a+b\) is called a binomial expression. The formula that gives the expansion of \((a+b)^{n}\) for any rational number \(n\), is called Binomial Theorem.

Q.2. What is meant by Binomial Theorem for Positive Integral Index?
Ans: If \(a, b \in R\) and \(n \in N\), then,
\((a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{0} a^{n-1} b+{ }^{n} C_{0} a^{n-2} b^{2}+\ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}\) is called Binomial Theorem for Positive Integral Index.

Q.3. What is \({ }^{n} C_{r}\) in Binomial Theorem?
Ans: Let \(n\) and \(r\) be non-negative integers, then \({ }^{n} C_{r}\) is called binomial coefficient
\({ }^{n} C_{r}=\frac{n !}{(n-r) ! r !}, \quad 0 \leq r \leq n\)

Q.4. What is the general formula for Binomial Theorem?
Ans: The formula for binomial theorem is given by,
\((a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{0} a^{n-1} b+{ }^{n} C_{0} a^{n-2} b^{2}+\ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}\)

Q.5. How do you interpret a binomial coefficient?
Ans: The binomial coefficients are the natural numbers that occur as coefficients in a binomial expansion. The values of binomial coefficient indicate how many times that particular term occurs in the entire expansion.

Learn about the Binomial Theorem here