Binomial theorem for any Index.
(1+x)n =
1+ nx + 
+ ..... + 
Observations
- Expansion is valid only when –1 < x <
1
- nCr can not be used because it is
defined only for natural number, so nCr will be written
as
- As the series never terminates, the number
of terms in the series is infinite. .
- General term of the series (1+x)-n
= Tr+1 = (-1)r
xr - General term of the series (1-x)-n
= Tr+1 = xr
- If first term is not 1, then make first term
unity in the following way:
- (a+ x)n = an(1+x/a)n
if
< 1
Important expansions
- (1+ x)-1 = 1- x +x2 –x3
+ . . . + (-1)rxr+. . .
- (1 - x)-1 = 1+ x +x2
+x3 + . . .+ xr + . . .
- (1+ x)-2 = 1- 2x +3x2
–4x3+ . . .+ (-1)r(r+1)xr+. . .
- (1 - x)-2 = 1+ 2x +3x2
+4x3 + . . .+ (r+1)xr+. . . .
- (1+x)-3 = 1- 3x +6x2
–10x3 +. . .+ (-1)r
xr+. . . - (1-x)-3 = 1+ 3x +6x2
+10x3 + . . .+ xr+.
. .
Multinomial Expansion
In the expansion of
(x1+x2 + . . . + xn)m where m, n ∈ N and x1, x2
, . . ., xn are independent variables, we have .
- Total number of
term in the expansion = m+n-1Cn-1
- Coefficient of
( where r1 +
r2 +. . . .+ rn = m, ri ∈N
is 
. - Sum of all the
coefficient is obtained by putting all the variables xi equal to1and
is nm.
Example.1
Find the total number of terms in the expansion of (1 +
a + b)10 and coefficient of a2b3. .
Solution
Total number of
terms = 10+3 –1C3-1 = 12C2 = 66
Coefficient of a2b3
= 
= 2520
Example.2
Find the number of
terms and coefficient of x5 in ( 1+x+x2)7.
Solution
Here the variables
1, x, x2 are not independent so the general formula is not
applicable but as power of x varries from 0 to 14, therefore total number of
terms = 15. .
Now x5
can be formed in three ways: 12 x5(x2)0,
13 x3( x2 )1 or
10 x1
(x2)2 , so total arrangement of coefficient
= 
=266
Alternative method:
coefficient of x5
in ( 1+ x+x2)7
= coefficient ox x5
in ( 1- x3)7( 1 - x)-7
= coefficient of x5
in ( 1 – 7x3+ . . )(1 + 7C1 x + 8C2
x2 + 9C3 x3 + . . ¥ ).
= 462 – 28 ´ 7 = 462 – 196 = 266.
