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=
nC
0 x
na
0 +
nC
1 x
n-1a
1 +
nC
2 x
n-2a
2 +………………+
nC
r x
n-ra
r …….+
nC
n x
0a
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=
nC
0 x
na
0 +
nC
1 x
n-1a
1 +
nC
2 x
n-2a
2 +………………+
nC
r x
n-ra
r …….+
nC
n x
0a
n
Step 1
Now the value of P(1)
(x+a)
1 =
1C
0 x
1a
0 +
1C
1 x
1-1a
1
=(x+a)
So P(1) is true
Step 2
Now the value of P(m)
(x+a)
m =
mC
0 x
ma
0 +
mC
1 x
m-1a
1 +
mC
2 x
m-2a
2 +………………+
mC
r x
m-ra
r …….+
mC
n x
0a
m
Now we have to prove
(x+a)
m+1 =
m+1C
0 x
m+1a
0 +
m+1C
1 x
ma
1 +
m+1C
2 x
m-1a
2 +………………+
m+1C
r x
m-r-1a
r …….+
m+1C
m+1 x
0a
m+1
Now
(x+a)
m+1 =(x+a)(x+a)
m
= (x+a)(
mC
0 x
ma
0 +
mC
1 x
m-1a
1 +
mC
2 x
m-2a
2 +………………+
mC
r x
m-ra
r …….+
mC
n x
0a
m)
=
mC
0 x
m+1a
0 + (
mC
1 +
mC
0) x
ma
1 + (
mC
2 +
mC
1) x
m-1a
2 + ………
+(
mC
m-1 +
mC
m) x
1a
m +
mC
m x
0a
m+1
As
mC
r-1 +
mC
r =
m+1C
r
So
=
m+1C
0 x
m+1a
0 +
m+1C
1 x
ma
1 +
m+1C
2 x
m-1a
2 +………………+
m+1C
r x
m-r-1a
r …….+
m+1C
m+1 x
0a
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
xn-kak
|
3
|
The coefficient nCr is each term is called binomial coefficient
|
4
|
(x-a)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.
|
(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
|
Examples: Expand using Binomial Theorem
(x2 +2)5
Solution: We know from binomial Theorem
(x+a)n= nC0 xna0 + nC1 xn-1a1 + nC2 xn-2a2 +………………+ nCr xn-rar …….+ nCn x0an
So putting values x=x2 ,a=2 and n=5
We get
(x2 +2)5 = 5C0 (x2)5 + 5C1 (x2)421 + 5C2 (x2)322 +5C3 (x2)223
+5C4 (x2)124 +5C5 (x2)025
=x10 + 20x8 +160x6 +640x4 +1280x2 +1024
General Term in Binomial Expansion
The First term would be =
nC
0 x
na
0
The Second term would be =
nC
1 x
n-1a
1
The Third term would be =
nC
2 x
n-2a
2
The Fourth term would be =
nC
3 x
n-3a
3
Like wise (k+1) term would be
T
k+1=
nC
k x
n-ka
k
This is called the general term also as every term can be find using this term
T
1= T
0+1=
nC
0 x
na
0
T
2= T
1+1=
nC
1 x
n-1a
1
Similarly for
T
k+1=
nC
k (-1)
k x
n-ka
k
Again similarly for
T
k+1=
nC
k x
k
Again similarly for
T
k+1=
nC
k (-1)
k x
k
To summarize it
Binomial term
|
(k+1) term
|
|
|
Tk+1= nCk xn-kak
|
|
|
Tk+1= nCk (-1)k xn-kak
|
|
|
Tk+1= nCk xk
|
|
Tk+1= nCk (-1)k xk
|
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=
nC
k x
k
Hence coefficient of (2k + 4)
th term will be
T
2k+4 = T
2k+3+1 =
24C
2k+3
and coefficient or (k - 2)
th term will be
T
k-2 = T
k-3+1 =
24C
k-3
As per question both the terms are equal
24C
2k+3 =
24C
k-3.
=> (2k + 3) + (k-3) = 24
.·. r = 8
Class 12 Maths
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