Binomial Theorem
An algebraic expression containing two terms is called binomial expression.
Example
( x+a)
(1/2 +x)
(2/x 1/x
^{3})
The general form of the binomial expression is (x+a) and expansion (x+a)
^{n} n? N is called Binomial expression
It was developed by Sir Issac newton
The general expression for the Binomial Theorem is
(x+a)
^{n}=
^{n}C
_{0} x
^{n}a
^{0} +
^{n}C
_{1} x
^{n1}a
^{1} +
^{n}C
_{2} x
^{n2}a
^{2} +………………+
^{n}C
_{r} x
^{nr}a
^{r} …….+
^{n}C
_{n} x
^{0}a
^{n}
Proof:
We can prove this theorem with the help of mathematical induction
Let us assume P(n) be the statement is
(x+a)
^{n}=
^{n}C
_{0} x
^{n}a
^{0} +
^{n}C
_{1} x
^{n1}a
^{1} +
^{n}C
_{2} x
^{n2}a
^{2} +………………+
^{n}C
_{r} x
^{nr}a
^{r} …….+
^{n}C
_{n} x
^{0}a
^{n}
Step 1
Now the value of P(1)
(x+a)
^{1} =
^{1}C
_{0} x
^{1}a
^{0} +
^{1}C
_{1} x
^{11}a
^{1}
=(x+a)
So P(1) is true
Step 2
Now the value of P(m)
(x+a)
^{m} =
^{m}C
_{0} x
^{m}a
^{0} +
^{m}C
_{1} x
^{m1}a
^{1} +
^{m}C
_{2} x
^{m2}a
^{2} +………………+
^{m}C
_{r} x
^{mr}a
^{r} …….+
^{m}C
_{n} x
^{0}a
^{m}
Now we have to prove
(x+a)
^{m+1} =
^{m+1}C
_{0} x
^{m+1}a
^{0} +
^{m+1}C
_{1} x
^{m}a
^{1} +
^{m+1}C
_{2} x
^{m1}a
^{2} +………………+
^{m+1}C
_{r} x
^{mr1}a
^{r} …….+
^{m+1}C
_{m+1} x
^{0}a
^{m+1}
Now
(x+a)
^{m+1} =(x+a)(x+a)
^{m}
= (x+a)(
^{m}C
_{0} x
^{m}a
^{0} +
^{m}C
_{1} x
^{m1}a
^{1} +
^{m}C
_{2} x
^{m2}a
^{2} +………………+
^{m}C
_{r} x
^{mr}a
^{r} …….+
^{m}C
_{n} x
^{0}a
^{m})
=
^{ m}C
_{0} x
^{m+1}a
^{0} + (
^{m}C
_{1} +
^{ m}C
_{0}) x
^{m}a
^{1} + (
^{m}C
_{2} +
^{ m}C
_{1}) x
^{m1}a
^{2} + ………
+(
^{m}C
_{m1} +
^{ m}C
_{m}) x
^{1}a
^{m} +
^{m}C
_{m} x
^{0}a
^{m+1}
As
^{m}C
_{r1} +
^{ m}C
_{r} =
^{m+1}C
_{r}
So
=
^{ m+1}C
_{0} x
^{m+1}a
^{0} +
^{m+1}C
_{1} x
^{m}a
^{1} +
^{m+1}C
_{2} x
^{m1}a
^{2} +………………+
^{m+1}C
_{r} x
^{mr1}a
^{r} …….+
^{m+1}C
_{m+1} x
^{0}a
^{m+1}
So by principle of Mathematical induction, P(n) is true for n? N
Important conclusion from Binomial Theorem
1

We can easily see that (x+a)^{n} has (n+1) terms as k can have values from 0 to n

2

The sum of indices of x and a in each is equal to n
x^{nk}a^{k}

3

The coefficient ^{n}C_{r} is each term is called binomial coefficient

4

(xa)^{n } can be treated as [x+(a)]^{n}
So
So the terms in the expansion are alternatively positive and negative. The last term is positive or negative depending on the values of n

5.

(1+x)^{n } can be treated as (x+a)^{n} where x=1 and a=x
So
This is the expansion is ascending order of power of x

6.

(1+x)^{n } can be treated as (x+a)^{n} where x=x and a=1
So
This is the expansion is descending order of power of x

7.

(1x)^{n } can be treated as (x+a)^{n} where x=1 and a=x
So
This is the expansion is ascending order of power of x

Examples: Expand using Binomial Theorem
(x^{2} +2)^{5}
Solution: We know from binomial Theorem
(x+a)^{n}= ^{n}C_{0} x^{n}a^{0} + ^{n}C_{1} x^{n1}a^{1} + ^{n}C_{2} x^{n2}a^{2} +………………+ ^{n}C_{r} x^{nr}a^{r} …….+ ^{n}C_{n} x^{0}a^{n}
So putting values x=x^{2} ,a=2 and n=5
We get
(x^{2} +2)^{5} =^{ 5}C_{0} (x^{2})^{5} + ^{5}C_{1} (x^{2})^{4}2^{1} + ^{5}C_{2} (x^{2})^{3}2^{2} +^{5}C_{3} (x^{2})^{2}2^{3}
+^{5}C_{4} (x^{2})^{1}2^{4} +^{5}C_{5} (x^{2})^{0}2^{5}
=x^{10} + 20x^{8} +160x^{6} +640x^{4} +1280x^{2} +1024
General Term in Binomial Expansion
The First term would be =
^{ n}C
_{0} x
^{n}a
^{0}
The Second term would be =
^{ n}C
_{1} x
^{n1}a
^{1}
The Third term would be =
^{ n}C
_{2} x
^{n2}a
^{2}
The Fourth term would be =
^{ n}C
_{3} x
^{n3}a
^{3}
Like wise (k+1) term would be
T
_{k+1}=
^{n}C
_{k} x
^{nk}a
^{k}
This is called the general term also as every term can be find using this term
T
_{1}= T
_{0+1}=
^{ n}C
_{0} x
^{n}a
^{0}
T
_{2}= T
_{1+1}=
^{ n}C
_{1} x
^{n1}a
^{1}
Similarly for
T
_{k+1}=
^{n}C
_{k} (1)
^{k} x
^{nk}a
^{k}
Again similarly for
T
_{k+1}=
^{n}C
_{k} x
^{k}
Again similarly for
T
_{k+1}=
^{n}C
_{k} (1)
^{k} x
^{k}
To summarize it
Binomial term

(k+1) term


T_{k+1}= ^{n}C_{k} x^{nk}a^{k}


T_{k+1}= ^{n}C_{k} (1)^{k} x^{nk}a^{k}


T_{k+1}= ^{n}C_{k} x^{k}


T_{k+1}= ^{n}C_{k} (1)^{k} x^{k}

Middle Term in Binomial Expansion
A binomial expansion contains (n+1) terms
If n is even then the middle term would [(n/2)+1 ]th term
If n is odd,then (n+1)/2 and (n+3)/3 are the middle term
Examples:
If the coefficient of (2k + 4) and (k  2) terms in the expansion of (1+x)
^{24} are equal then find the value of k
Solution:
The general term of (1 + x)
^{n} is
T
_{k+1}=
^{n}C
_{k} x
^{k}
Hence coefficient of (2k + 4)
^{th} term will be
T
_{2k+4} = T
_{2k+3+1} =
^{24}C
_{2k+3}
and coefficient or (k  2)
^{th} term will be
T
_{k2} = T
_{k3+1} =
^{24}C
_{k3}
As per question both the terms are equal
^{24}C
_{2k+3} =
^{24}C
_{k3}.
=> (2k + 3) + (k3) = 24
.·. r = 8
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Class 12 Physics