Page 1
126 MATHEMATICS
vMathematics is a most exact science and its conclusions are capable of
absolute proofs. – C.P. STEINMETZv
7.1 Introduction
In earlier classes, we have learnt how to find the squares
and cubes of binomials like a + b and a – b. Using them, we
could evaluate the numerical values of numbers like
(98)
2
= (100 – 2)
2
, (999)
3
= (1000 – 1)
3
, etc. However, for
higher powers like (98)
5
, (101)
6
, etc., the calculations become
difficult by using repeated multiplication. This difficulty was
overcome by a theorem known as binomial theorem. It gives
an easier way to expand (a + b)
n
, where n is an integer or a
rational number. In this Chapter, we study binomial theorem
for positive integral indices only.
7.2 Binomial Theorem for Positive Integral Indices
Let us have a look at the following identities done earlier:
(a+ b)
0
= 1 a + b ? 0
(a+ b)
1
= a + b
(a+ b)
2
= a
2
+ 2ab + b
2
(a+ b)
3
= a
3
+ 3a
2
b + 3ab
2
+ b
3
(a+ b)
4
= (a + b)
3
(a + b) = a
4
+ 4a
3
b + 6a
2
b
2
+ 4ab
3
+ b
4
In these expansions, we observe that
(i) The total number of terms in the expansion is one 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.
(ii) Powers of the first quantity ‘a’ go on decreasing by 1 whereas the powers of the
second quantity ‘b’ increase by 1, in the successive terms.
(iii) In each term of the expansion, the sum of the indices of a and b is the same and
is equal to the index of a + b.
7 Chapter
Blaise Pascal
(16231662)
BINOMIAL THEOREM
202425
Page 2
126 MATHEMATICS
vMathematics is a most exact science and its conclusions are capable of
absolute proofs. – C.P. STEINMETZv
7.1 Introduction
In earlier classes, we have learnt how to find the squares
and cubes of binomials like a + b and a – b. Using them, we
could evaluate the numerical values of numbers like
(98)
2
= (100 – 2)
2
, (999)
3
= (1000 – 1)
3
, etc. However, for
higher powers like (98)
5
, (101)
6
, etc., the calculations become
difficult by using repeated multiplication. This difficulty was
overcome by a theorem known as binomial theorem. It gives
an easier way to expand (a + b)
n
, where n is an integer or a
rational number. In this Chapter, we study binomial theorem
for positive integral indices only.
7.2 Binomial Theorem for Positive Integral Indices
Let us have a look at the following identities done earlier:
(a+ b)
0
= 1 a + b ? 0
(a+ b)
1
= a + b
(a+ b)
2
= a
2
+ 2ab + b
2
(a+ b)
3
= a
3
+ 3a
2
b + 3ab
2
+ b
3
(a+ b)
4
= (a + b)
3
(a + b) = a
4
+ 4a
3
b + 6a
2
b
2
+ 4ab
3
+ b
4
In these expansions, we observe that
(i) The total number of terms in the expansion is one 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.
(ii) Powers of the first quantity ‘a’ go on decreasing by 1 whereas the powers of the
second quantity ‘b’ increase by 1, in the successive terms.
(iii) In each term of the expansion, the sum of the indices of a and b is the same and
is equal to the index of a + b.
7 Chapter
Blaise Pascal
(16231662)
BINOMIAL THEOREM
202425
BINOMIAL THEOREM 127
We now arrange the coefficients in these expansions as follows (Fig 7.1):
Do we observe any pattern in this table that will help us to write the next row? Y es we
do. It can be seen that the addition of 1’s in the row for index 1 gives rise to 2 in the row
for index 2. The addition of 1, 2 and 2, 1 in the row for index 2, gives rise to 3 and 3 in
the row for index 3 and so on. Also, 1 is present at the beginning and at the end of each
row. This can be continued till any index of our interest.
We can extend the pattern given in Fig 7.2 by writing a few more rows.
Pascal’s Triangle
The structure given in Fig 7.2 looks like a triangle with 1 at the top vertex and running
down the two slanting sides. This array of numbers is known as Pascal’s triangle,
after the name of French mathematician Blaise Pascal. It is also known as Meru
Prastara by Pingla.
Expansions for the higher powers of a binomial are also possible by using Pascal’s
triangle. Let us expand (2x + 3y)
5
by using Pascal’s triangle. The row for index 5 is
1 5 10 10 5 1
Using this row and our observations (i), (ii) and (iii), we get
(2x + 3y)
5
= (2x)
5
+ 5(2x)
4
(3y) + 10(2x)
3
(3y)
2
+10 (2x)
2
(3y)
3
+ 5(2x)(3y)
4
+(3y)
5
= 32x
5
+ 240x
4
y + 720x
3
y
2
+ 1080x
2
y
3
+ 810xy
4
+ 243y
5
.
Fig 7.1
Fig 7.2
202425
Page 3
126 MATHEMATICS
vMathematics is a most exact science and its conclusions are capable of
absolute proofs. – C.P. STEINMETZv
7.1 Introduction
In earlier classes, we have learnt how to find the squares
and cubes of binomials like a + b and a – b. Using them, we
could evaluate the numerical values of numbers like
(98)
2
= (100 – 2)
2
, (999)
3
= (1000 – 1)
3
, etc. However, for
higher powers like (98)
5
, (101)
6
, etc., the calculations become
difficult by using repeated multiplication. This difficulty was
overcome by a theorem known as binomial theorem. It gives
an easier way to expand (a + b)
n
, where n is an integer or a
rational number. In this Chapter, we study binomial theorem
for positive integral indices only.
7.2 Binomial Theorem for Positive Integral Indices
Let us have a look at the following identities done earlier:
(a+ b)
0
= 1 a + b ? 0
(a+ b)
1
= a + b
(a+ b)
2
= a
2
+ 2ab + b
2
(a+ b)
3
= a
3
+ 3a
2
b + 3ab
2
+ b
3
(a+ b)
4
= (a + b)
3
(a + b) = a
4
+ 4a
3
b + 6a
2
b
2
+ 4ab
3
+ b
4
In these expansions, we observe that
(i) The total number of terms in the expansion is one 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.
(ii) Powers of the first quantity ‘a’ go on decreasing by 1 whereas the powers of the
second quantity ‘b’ increase by 1, in the successive terms.
(iii) In each term of the expansion, the sum of the indices of a and b is the same and
is equal to the index of a + b.
7 Chapter
Blaise Pascal
(16231662)
BINOMIAL THEOREM
202425
BINOMIAL THEOREM 127
We now arrange the coefficients in these expansions as follows (Fig 7.1):
Do we observe any pattern in this table that will help us to write the next row? Y es we
do. It can be seen that the addition of 1’s in the row for index 1 gives rise to 2 in the row
for index 2. The addition of 1, 2 and 2, 1 in the row for index 2, gives rise to 3 and 3 in
the row for index 3 and so on. Also, 1 is present at the beginning and at the end of each
row. This can be continued till any index of our interest.
We can extend the pattern given in Fig 7.2 by writing a few more rows.
Pascal’s Triangle
The structure given in Fig 7.2 looks like a triangle with 1 at the top vertex and running
down the two slanting sides. This array of numbers is known as Pascal’s triangle,
after the name of French mathematician Blaise Pascal. It is also known as Meru
Prastara by Pingla.
Expansions for the higher powers of a binomial are also possible by using Pascal’s
triangle. Let us expand (2x + 3y)
5
by using Pascal’s triangle. The row for index 5 is
1 5 10 10 5 1
Using this row and our observations (i), (ii) and (iii), we get
(2x + 3y)
5
= (2x)
5
+ 5(2x)
4
(3y) + 10(2x)
3
(3y)
2
+10 (2x)
2
(3y)
3
+ 5(2x)(3y)
4
+(3y)
5
= 32x
5
+ 240x
4
y + 720x
3
y
2
+ 1080x
2
y
3
+ 810xy
4
+ 243y
5
.
Fig 7.1
Fig 7.2
202425
128 MATHEMATICS
Now, if we want to find the expansion of (2x + 3y)
12
, we are first required to get
the row for index 12. This can be done by writing all the rows of the Pascal’s triangle
till index 12. This is a slightly lengthy process. The process, as you observe, will become
more difficult, if we need the expansions involving still larger powers.
W e thus try to find a rule that will help us to find the expansion of the binomial for
any power without writing all the rows of the Pascal’s triangle, that come before the
row of the desired index.
For this, we make use of the concept of combinations studied earlier to rewrite
the numbers in the Pascal’s triangle. We know that
!
C
!( )!
n
r
n
r n – r
=
, 0 = r = n and
n is a nonnegative integer. Also,
n
C
0
= 1 =
n
C
n
The Pascal’s triangle can now be rewritten as (Fig 7.3)
Observing this pattern, we can now write the row of the Pascal’s triangle for any index
without writing the earlier rows. For example, for the index 7 the row would be
7
C
0
7
C
1
7
C
2
7
C
3
7
C
4
7
C
5
7
C
6
7
C
7.
Thus, using this row and the observations (i), (ii) and (iii), we have
(a + b)
7
=
7
C
0
a
7
+ 7C
1
a
6
b +
7
C
2
a
5
b
2
+
7
C
3
a
4
b
3
+ 7C
4
a
3
b
4
+
7
C
5
a
2
b
5
+
7
C
6
ab
6
+
7
C
7
b
7
An expansion of a binomial to any positive integral index say n can now be visualised
using these observations. W e are now in a position to write the expansion of a binomial
to any positive integral index.
Fig 7.3 Pascal’s triangle
202425
Page 4
126 MATHEMATICS
vMathematics is a most exact science and its conclusions are capable of
absolute proofs. – C.P. STEINMETZv
7.1 Introduction
In earlier classes, we have learnt how to find the squares
and cubes of binomials like a + b and a – b. Using them, we
could evaluate the numerical values of numbers like
(98)
2
= (100 – 2)
2
, (999)
3
= (1000 – 1)
3
, etc. However, for
higher powers like (98)
5
, (101)
6
, etc., the calculations become
difficult by using repeated multiplication. This difficulty was
overcome by a theorem known as binomial theorem. It gives
an easier way to expand (a + b)
n
, where n is an integer or a
rational number. In this Chapter, we study binomial theorem
for positive integral indices only.
7.2 Binomial Theorem for Positive Integral Indices
Let us have a look at the following identities done earlier:
(a+ b)
0
= 1 a + b ? 0
(a+ b)
1
= a + b
(a+ b)
2
= a
2
+ 2ab + b
2
(a+ b)
3
= a
3
+ 3a
2
b + 3ab
2
+ b
3
(a+ b)
4
= (a + b)
3
(a + b) = a
4
+ 4a
3
b + 6a
2
b
2
+ 4ab
3
+ b
4
In these expansions, we observe that
(i) The total number of terms in the expansion is one 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.
(ii) Powers of the first quantity ‘a’ go on decreasing by 1 whereas the powers of the
second quantity ‘b’ increase by 1, in the successive terms.
(iii) In each term of the expansion, the sum of the indices of a and b is the same and
is equal to the index of a + b.
7 Chapter
Blaise Pascal
(16231662)
BINOMIAL THEOREM
202425
BINOMIAL THEOREM 127
We now arrange the coefficients in these expansions as follows (Fig 7.1):
Do we observe any pattern in this table that will help us to write the next row? Y es we
do. It can be seen that the addition of 1’s in the row for index 1 gives rise to 2 in the row
for index 2. The addition of 1, 2 and 2, 1 in the row for index 2, gives rise to 3 and 3 in
the row for index 3 and so on. Also, 1 is present at the beginning and at the end of each
row. This can be continued till any index of our interest.
We can extend the pattern given in Fig 7.2 by writing a few more rows.
Pascal’s Triangle
The structure given in Fig 7.2 looks like a triangle with 1 at the top vertex and running
down the two slanting sides. This array of numbers is known as Pascal’s triangle,
after the name of French mathematician Blaise Pascal. It is also known as Meru
Prastara by Pingla.
Expansions for the higher powers of a binomial are also possible by using Pascal’s
triangle. Let us expand (2x + 3y)
5
by using Pascal’s triangle. The row for index 5 is
1 5 10 10 5 1
Using this row and our observations (i), (ii) and (iii), we get
(2x + 3y)
5
= (2x)
5
+ 5(2x)
4
(3y) + 10(2x)
3
(3y)
2
+10 (2x)
2
(3y)
3
+ 5(2x)(3y)
4
+(3y)
5
= 32x
5
+ 240x
4
y + 720x
3
y
2
+ 1080x
2
y
3
+ 810xy
4
+ 243y
5
.
Fig 7.1
Fig 7.2
202425
128 MATHEMATICS
Now, if we want to find the expansion of (2x + 3y)
12
, we are first required to get
the row for index 12. This can be done by writing all the rows of the Pascal’s triangle
till index 12. This is a slightly lengthy process. The process, as you observe, will become
more difficult, if we need the expansions involving still larger powers.
W e thus try to find a rule that will help us to find the expansion of the binomial for
any power without writing all the rows of the Pascal’s triangle, that come before the
row of the desired index.
For this, we make use of the concept of combinations studied earlier to rewrite
the numbers in the Pascal’s triangle. We know that
!
C
!( )!
n
r
n
r n – r
=
, 0 = r = n and
n is a nonnegative integer. Also,
n
C
0
= 1 =
n
C
n
The Pascal’s triangle can now be rewritten as (Fig 7.3)
Observing this pattern, we can now write the row of the Pascal’s triangle for any index
without writing the earlier rows. For example, for the index 7 the row would be
7
C
0
7
C
1
7
C
2
7
C
3
7
C
4
7
C
5
7
C
6
7
C
7.
Thus, using this row and the observations (i), (ii) and (iii), we have
(a + b)
7
=
7
C
0
a
7
+ 7C
1
a
6
b +
7
C
2
a
5
b
2
+
7
C
3
a
4
b
3
+ 7C
4
a
3
b
4
+
7
C
5
a
2
b
5
+
7
C
6
ab
6
+
7
C
7
b
7
An expansion of a binomial to any positive integral index say n can now be visualised
using these observations. W e are now in a position to write the expansion of a binomial
to any positive integral index.
Fig 7.3 Pascal’s triangle
202425
BINOMIAL THEOREM 129
7.2.1 Binomial theorem for any positive integer n,
(a + b)
n
=
n
C
0
a
n
+
n
C
1
a
n–1
b +
n
C
2
a
n–2
b
2
+ ...+
n
C
n – 1
a.b
n–1
+
n
C
n
b
n
Proof The proof is obtained by applying principle of mathematical induction.
Let the given statement be
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
+ ...+
n
C
n–1
a.b
n – 1
+
n
C
n
b
n
For n = 1, we have
P (1) : (a + b)
1
=
1
C
0
a
1
+
1
C
1
b
1
= a + b
Thus, P (1) is true.
Suppose P (k) is true for some positive integer k, i.e.
(a + b)
k
=
k
C
0
a
k
+
k
C
1
a
k – 1
b +
k
C
2
a
k – 2
b
2
+ ...+
k
C
k
b
k
...
(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
+ ...+
k + 1
C
k+1
b
k + 1
Now, (a + b)
k + 1
= (a + b) (a + b)
k
= (a + b) (
k
C
0
a
k
+
k
C
1
a
k – 1
b +
k
C
2
a
k – 2
b
2
+...+
k
C
k – 1
ab
k – 1
+
k
C
k
b
k
)
[from (1)]
=
k
C
0
a
k + 1
+
k
C
1
a
k
b +
k
C
2
a
k – 1
b
2
+...+
k
C
k – 1
a
2
b
k – 1
+
k
C
k
ab
k
+
k
C
0
a
k
b
+
k
C
1
a
k – 1
b
2
+
k
C
2
a
k – 2
b
3
+...+
k
C
k1
ab
k
+
k
C
k
b
k + 1
[by actual multiplication]
=
k
C
0
a
k + 1
+ (
k
C
1
+
k
C
0
) a
k
b + (
k
C
2
+
k
C
1
)a
k – 1
b
2
+ ...
+ (
k
C
k
+
k
C
k–1
) ab
k
+
k
C
k
b
k + 1
[grouping like terms]
=
k + 1
C
0
a
k + 1
+
k + 1
C
1
a
k
b +
k + 1
C
2
a
k – 1
b
2
+...+
k + 1
C
k
ab
k
+
k + 1
C
k + 1
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
)
Thus, it has been 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.
We illustrate this theorem by expanding (x + 2)
6
:
(x + 2)
6
=
6
C
0
x
6
+
6
C
1
x
5
.2 +
6
C
2
x
4
2
2
+
6
C
3
x
3
.2
3
+
6
C
4
x
2
.2
4
+
6
C
5
x.2
5
+
6
C
6
.2
6
.
= x
6
+ 12x
5
+ 60x
4
+ 160x
3
+
240x
2
+ 192x + 64
Thus (x + 2)
6
= x
6
+ 12x
5
+ 60x
4
+ 160x
3
+
240x
2
+ 192x + 64.
202425
Page 5
126 MATHEMATICS
vMathematics is a most exact science and its conclusions are capable of
absolute proofs. – C.P. STEINMETZv
7.1 Introduction
In earlier classes, we have learnt how to find the squares
and cubes of binomials like a + b and a – b. Using them, we
could evaluate the numerical values of numbers like
(98)
2
= (100 – 2)
2
, (999)
3
= (1000 – 1)
3
, etc. However, for
higher powers like (98)
5
, (101)
6
, etc., the calculations become
difficult by using repeated multiplication. This difficulty was
overcome by a theorem known as binomial theorem. It gives
an easier way to expand (a + b)
n
, where n is an integer or a
rational number. In this Chapter, we study binomial theorem
for positive integral indices only.
7.2 Binomial Theorem for Positive Integral Indices
Let us have a look at the following identities done earlier:
(a+ b)
0
= 1 a + b ? 0
(a+ b)
1
= a + b
(a+ b)
2
= a
2
+ 2ab + b
2
(a+ b)
3
= a
3
+ 3a
2
b + 3ab
2
+ b
3
(a+ b)
4
= (a + b)
3
(a + b) = a
4
+ 4a
3
b + 6a
2
b
2
+ 4ab
3
+ b
4
In these expansions, we observe that
(i) The total number of terms in the expansion is one 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.
(ii) Powers of the first quantity ‘a’ go on decreasing by 1 whereas the powers of the
second quantity ‘b’ increase by 1, in the successive terms.
(iii) In each term of the expansion, the sum of the indices of a and b is the same and
is equal to the index of a + b.
7 Chapter
Blaise Pascal
(16231662)
BINOMIAL THEOREM
202425
BINOMIAL THEOREM 127
We now arrange the coefficients in these expansions as follows (Fig 7.1):
Do we observe any pattern in this table that will help us to write the next row? Y es we
do. It can be seen that the addition of 1’s in the row for index 1 gives rise to 2 in the row
for index 2. The addition of 1, 2 and 2, 1 in the row for index 2, gives rise to 3 and 3 in
the row for index 3 and so on. Also, 1 is present at the beginning and at the end of each
row. This can be continued till any index of our interest.
We can extend the pattern given in Fig 7.2 by writing a few more rows.
Pascal’s Triangle
The structure given in Fig 7.2 looks like a triangle with 1 at the top vertex and running
down the two slanting sides. This array of numbers is known as Pascal’s triangle,
after the name of French mathematician Blaise Pascal. It is also known as Meru
Prastara by Pingla.
Expansions for the higher powers of a binomial are also possible by using Pascal’s
triangle. Let us expand (2x + 3y)
5
by using Pascal’s triangle. The row for index 5 is
1 5 10 10 5 1
Using this row and our observations (i), (ii) and (iii), we get
(2x + 3y)
5
= (2x)
5
+ 5(2x)
4
(3y) + 10(2x)
3
(3y)
2
+10 (2x)
2
(3y)
3
+ 5(2x)(3y)
4
+(3y)
5
= 32x
5
+ 240x
4
y + 720x
3
y
2
+ 1080x
2
y
3
+ 810xy
4
+ 243y
5
.
Fig 7.1
Fig 7.2
202425
128 MATHEMATICS
Now, if we want to find the expansion of (2x + 3y)
12
, we are first required to get
the row for index 12. This can be done by writing all the rows of the Pascal’s triangle
till index 12. This is a slightly lengthy process. The process, as you observe, will become
more difficult, if we need the expansions involving still larger powers.
W e thus try to find a rule that will help us to find the expansion of the binomial for
any power without writing all the rows of the Pascal’s triangle, that come before the
row of the desired index.
For this, we make use of the concept of combinations studied earlier to rewrite
the numbers in the Pascal’s triangle. We know that
!
C
!( )!
n
r
n
r n – r
=
, 0 = r = n and
n is a nonnegative integer. Also,
n
C
0
= 1 =
n
C
n
The Pascal’s triangle can now be rewritten as (Fig 7.3)
Observing this pattern, we can now write the row of the Pascal’s triangle for any index
without writing the earlier rows. For example, for the index 7 the row would be
7
C
0
7
C
1
7
C
2
7
C
3
7
C
4
7
C
5
7
C
6
7
C
7.
Thus, using this row and the observations (i), (ii) and (iii), we have
(a + b)
7
=
7
C
0
a
7
+ 7C
1
a
6
b +
7
C
2
a
5
b
2
+
7
C
3
a
4
b
3
+ 7C
4
a
3
b
4
+
7
C
5
a
2
b
5
+
7
C
6
ab
6
+
7
C
7
b
7
An expansion of a binomial to any positive integral index say n can now be visualised
using these observations. W e are now in a position to write the expansion of a binomial
to any positive integral index.
Fig 7.3 Pascal’s triangle
202425
BINOMIAL THEOREM 129
7.2.1 Binomial theorem for any positive integer n,
(a + b)
n
=
n
C
0
a
n
+
n
C
1
a
n–1
b +
n
C
2
a
n–2
b
2
+ ...+
n
C
n – 1
a.b
n–1
+
n
C
n
b
n
Proof The proof is obtained by applying principle of mathematical induction.
Let the given statement be
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
+ ...+
n
C
n–1
a.b
n – 1
+
n
C
n
b
n
For n = 1, we have
P (1) : (a + b)
1
=
1
C
0
a
1
+
1
C
1
b
1
= a + b
Thus, P (1) is true.
Suppose P (k) is true for some positive integer k, i.e.
(a + b)
k
=
k
C
0
a
k
+
k
C
1
a
k – 1
b +
k
C
2
a
k – 2
b
2
+ ...+
k
C
k
b
k
...
(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
+ ...+
k + 1
C
k+1
b
k + 1
Now, (a + b)
k + 1
= (a + b) (a + b)
k
= (a + b) (
k
C
0
a
k
+
k
C
1
a
k – 1
b +
k
C
2
a
k – 2
b
2
+...+
k
C
k – 1
ab
k – 1
+
k
C
k
b
k
)
[from (1)]
=
k
C
0
a
k + 1
+
k
C
1
a
k
b +
k
C
2
a
k – 1
b
2
+...+
k
C
k – 1
a
2
b
k – 1
+
k
C
k
ab
k
+
k
C
0
a
k
b
+
k
C
1
a
k – 1
b
2
+
k
C
2
a
k – 2
b
3
+...+
k
C
k1
ab
k
+
k
C
k
b
k + 1
[by actual multiplication]
=
k
C
0
a
k + 1
+ (
k
C
1
+
k
C
0
) a
k
b + (
k
C
2
+
k
C
1
)a
k – 1
b
2
+ ...
+ (
k
C
k
+
k
C
k–1
) ab
k
+
k
C
k
b
k + 1
[grouping like terms]
=
k + 1
C
0
a
k + 1
+
k + 1
C
1
a
k
b +
k + 1
C
2
a
k – 1
b
2
+...+
k + 1
C
k
ab
k
+
k + 1
C
k + 1
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
)
Thus, it has been 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.
We illustrate this theorem by expanding (x + 2)
6
:
(x + 2)
6
=
6
C
0
x
6
+
6
C
1
x
5
.2 +
6
C
2
x
4
2
2
+
6
C
3
x
3
.2
3
+
6
C
4
x
2
.2
4
+
6
C
5
x.2
5
+
6
C
6
.2
6
.
= x
6
+ 12x
5
+ 60x
4
+ 160x
3
+
240x
2
+ 192x + 64
Thus (x + 2)
6
= x
6
+ 12x
5
+ 60x
4
+ 160x
3
+
240x
2
+ 192x + 64.
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130 MATHEMATICS
Observations
1. The notation ?
=

n
k
k k n
k
n
b a
0
C
stands for
n
C
0
a
n
b
0
+
n
C
1
a
n–1
b
1
+ ...+
n
C
r
a
n–r
b
r
+ ...+
n
C
n
a
n–n
b
n
, where b
0
= 1 = a
n–n
.
Hence the theorem can also be stated as
?
=

= +
n
k
k k n
k
n n
b a b a
0
C ) (
.
2. The coefficients
n
C
r
occuring 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.
7.2.2 Some special cases In the expansion of (a + b)
n
,
(i) Taking a = x and b = – y, we obtain
(x – y)
n
= [x + (–y)]
n
=
n
C
0
x
n
+
n
C
1
x
n – 1
(–y) +
n
C
2
x
n–2
(–y)
2
+
n
C
3
x
n–3
(–y)
3
+ ... +
n
C
n
(–y)
n
=
n
C
0
x
n
–
n
C
1
x
n – 1
y +
n
C
2
x
n – 2
y
2
–
n
C
3
x
n – 3
y
3
+ ... + (–1)
n
n
C
n
y
n
Thus (x–y)
n
=
n
C
0
x
n
–
n
C
1
x
n – 1
y +
n
C
2
x
n – 2
y
2
+ ... + (–1)
n
n
C
n
y
n
Using this, we have (x–2y)
5
=
5
C
0
x
5
–
5
C
1
x
4
(2y) +
5
C
2
x
3
(2y)
2
–
5
C
3
x
2
(2y)
3
+
5
C
4
x(2y)
4
–
5
C
5
(2y)
5
= x
5
–10x
4
y + 40x
3
y
2
– 80x
2
y
3
+ 80xy
4
– 32y
5
.
(ii) Taking a = 1, b = x, we obtain
(1 + x)
n
=
n
C
0
(1)
n
+
n
C
1
(1)
n – 1
x +
n
C
2
(1)
n – 2
x
2
+ ... +
n
C
n
x
n
=
n
C
0
+
n
C
1
x +
n
C
2
x
2
+
n
C
3
x
3
+ ... +
n
C
n
x
n
Thus (1 + x)
n
=
n
C
0
+
n
C
1
x +
n
C
2
x
2
+
n
C
3
x
3
+ ... +
n
C
n
x
n
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