Binomial Theorem

Binomial Expression

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

Binomial Theorem
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 :

Similarly,

(a−x)n = nC0annC1an−1 x + nC2 an-2 x2 − . . . +( −1)r nCr an−r xr + . . . + ( −1)n nCnxn

Illustration : Expand

Solution:

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

Solution:

General term ((r + 1) th term) in

15Cr(x2)15−r(3a/x)r

= 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

Note:

∎ 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)128128C1(290)127 + ……− 128C125(290)3 + 128C126(290)2128C127(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

Next Page » : Middle Terms

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