Important Binomial Theorem & its Simple Applications Formulas for JEE and NEET

 Page 1


Page # 26
BINOMIAL THEOREM
1. Statement of Binomial theorem :  If a, b ? R and n ? N, then
(a + b)
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
r
 a
n–r
 b
r
 +...+ 
n
C
n
 a
0
 b
n
= ?
?
?
n
0 r
r r n
r
n
b a C
2. Properties of Binomial Theorem :
(i) General term :  T
r+1
 = 
n
C
r
 a
n–r
 b
r
(ii) Middle term (s) :
(a) If n is even, there is only one middle term,
which is 
?
?
?
?
?
? ?
2
2 n
th term.
(b) If n is odd, there are two middle terms,
which are 
?
?
?
?
?
? ?
2
1 n
th and 
?
?
?
?
?
?
?
?
1
2
1 n
th terms.
Page 2


Page # 26
BINOMIAL THEOREM
1. Statement of Binomial theorem :  If a, b ? R and n ? N, then
(a + b)
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
r
 a
n–r
 b
r
 +...+ 
n
C
n
 a
0
 b
n
= ?
?
?
n
0 r
r r n
r
n
b a C
2. Properties of Binomial Theorem :
(i) General term :  T
r+1
 = 
n
C
r
 a
n–r
 b
r
(ii) Middle term (s) :
(a) If n is even, there is only one middle term,
which is 
?
?
?
?
?
? ?
2
2 n
th term.
(b) If n is odd, there are two middle terms,
which are 
?
?
?
?
?
? ?
2
1 n
th and 
?
?
?
?
?
?
?
?
1
2
1 n
th terms.
Page # 27
3. Multinomial Theorem :
(x
1
 + x
2
 + x
3
 + ........... x
k
)
n
= 
?
???? n r ... r r
k 2 1
k 2 1
! r !... r ! r
! n
 
k 2 1
r
k
r
2
r
1
x ... x . x
Here total number of terms in the expansion =  
n+k–1
C
k–1
4. Application of Binomial Theorem :
If 
n
) B A ( ? = ? + f where ? and  n are positive integers, n being odd and
0 < f < 1  then ( ? + f) f = k
n
 where A – B
2
 = k > 0 and A – B < 1.
If n is an even integer, then ( ? + f) (1 – f) = k
n
5. Properties of Binomial Coefficients :
(i)
n
C
0
 + 
n
C
1
 + 
n
C
2
 + ........+ 
n
C
n
 = 2
n
(ii)
n
C
0
 – 
n
C
1
 + 
n
C
2
 – 
n
C
3
 + ............. + (–1)
n
 
n
C
n
  = 0
(iii)
n
C
0
 + 
n
C
2
 + 
n
C
4
 + .... = 
n
C
1
 + 
n
C
3
 + 
n
C
5
 + .... = 2
n–1
(iv)
n
C
r
 + 
n
C
r–1
  = 
n+1
C
r
(v)
1 r
n
r
n
C
C
?
 = 
r
1 r n ? ?
6. Binomial Theorem For Negative Integer Or Fractional Indices
(1 + x)
n
 = 1 + nx + 
! 2
) 1 n ( n ?
 x
2
 + 
! 3
) 2 n )( 1 n ( n ? ?
 x
3
 + .... +
! r
) 1 r n ).......( 2 n )( 1 n ( n ? ? ? ?
 x
r
 + ....,| x | < 1.
 T
r+1
 = 
! r
) 1 r n ( )......... 2 n )( 1 n ( n ? ? ? ?
 x
r