Any algebraic expression consisting of only two terms is known as a binomial expression.
It’s expansion in power of x is known as the binomial expansion.
e.g. (i) a + x (ii) a2 + 1/x2 (iii) 4x − 6y
Such formula by which any power of a binomial expression can be expanded in the form of a series is known as binomial theorem.
For a positive integer n , the expansion is given by :
(a + x)n = nC0an + nC1an−1 x + nC2 an−2 x2 + . . . + nCr an−r xr + . . . + nCnxn
where nC0, nC1 , nC2 , . . . , nCn are called Binomial co-efficients and there are (n + 1) terms in the expansion.
The value of nCr is defined as :
(a−x)n = nC0an − nC1an−1 x + nC2 an-2 x2 − . . . +( −1)r nCr an−r xr + . . . + ( −1)n nCnxn
Illustration : Expand
7C0x7 + 7C1x6 + 7C2x5 + 7C3x4 + 7C4x3 + 7C5x2 + 7C6 x + 7C7
= x7 + 7x5 + 21x3 + 35 x + 35/x + 21/x3 + 7/x5 + 1/x7
General Term in the Expansion:
The general term in the expansion of (a + x)n is the (r + 1)th term given as :
tr+1 = nCr an-r xr .
Similarly the general term in the expansion of (x + a)n is given as :
tr+1 = nCr xn-r ar .
The terms are considered from the beginning.
The (r + 1)th term from the end = (n − r + 1)th term from the beginning.
Corollary: Coefficient of xr in expansion of (1 + x)n is nCr
Illustration : Find the co-efficient of x24 in
General term ((r + 1) th term) in
= 15Cr x30-2r(3rar/xr)
= 15Cr 3rar x30-3r
If this term contains x24 .
Then 30−3r = 24
=> 3r = 6 => r = 2
Therefore, the co-efficient of x24 = 15 C2 . 9a2
Coefficients of Equidistant Terms from Beginning & end
The binomial coefficients in the expansion of (a + x)n equidistant from the beginning and the end are equal.
i.e. nCr = nCn−r
∎ If nCx = nCy , then either x = y or x + y = n
Illustration 3: If the co-efficient of (2r + 4)th term and (r − 2)th term in the expansion of (1 + x)18are equal, find r
Solution: Since, co-efficient of (2r + 4)th term in (1 + x)18 = 18C2r + 3
Co-efficient of (r − 2) th term = 18Cr − 3
=> 18C2r + 3 = 18Cr − 3
=> 2r + 3 + r − 3 = 18 => 3r = 18
=> r = 6
Illustration : Find the last three digits of 17256
Solution: We have 172 = 289 =290 − 1
Now, 17256 = (172)128 = (290 − 1)128
= 128C0(290)128 − 128C1(290)127 + ……− 128C125(290)3 + 128C126(290)2 − 128C127(290) + 1
, where m is a positive integer.
= 1000 m + (128) (290) [(127) (145) − 1 ] + 1 = 1000 m + (128) (290) (18414) + 1
= 1000 [m + 683527] + 680 + 1 = 1000 [m + 683527] + 681
Thus, the last three digits of 17256 are 681