JEE-Mathematics
Page No: 1
BINOMIAL THEOREM 1.
BINOMIAL EXPRESSION : Any algebraic expression which contains two dissimilar terms is called binomial expression. For example : x – y, xy +
2.
1 1 1 , 1, 3 etc. x z (x y )1 / 3
BINOMIAL THEOREM : The formula by which any positive integral power of a binomial expression can be expanded in the form of a series is known as BINOMIAL THEOREM. n
If x, y R and n N, then : ( x + y)n = nC0xn + nC1xn-1 y + nC2xn-2 y2 + ..... + nCrxn-r yr + ..... + nCnyn = n C r x n r y r r 0
This theorem can be proved by induction. Observations : (a) The number of terms in the expansion is ( n+1) i.e. one more than the index. (b) The sum of the indices of x & y in each term is n. (c) The binomial coefficients of the terms (nC0, nC1.....) equidistant from the beginning and the end are equal. i.e. n C r = n C r –1 (d)
n Symbol nCr can also be denoted by , C(n, r) or A nr . r
Some important expansions : (i)
(1 + x) n = n C 0 + n C 1 x + n C 2 x 2 + ........ + nC nx n .
(ii)
(1 – x) n = n C 0 – nC 1x + n C 2 x 2 + ........ + (–1) n . n C n x n .
Note :
The coefficient of x r in (1+x) n = nC r & that in (1–x) n = (–1) r . nC r
Illustration 1 :
Expand : (y + 2)6.
Solution :
6C y6 + 6C y5.2 + 6C y4.22 + 6C y3.23 + 6C y2. 24 + 6C y1 . 25 + 0 1 2 3 4 5
6C
6 6 . 2 .
= y6 + 12y5 + 60y4 + 160y3 + 240y2 + 192y + 64.
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Illustration 2 :
2y 2 Write first 4 terms of 1 5
Solution :
2y 2 7 2y 2 7 2y 2 7C , 7C C , C 3 , 2 0 1 5 5 5
Illustration 3 :
The value of
2
18 7 3.18.7.25 3
3
is -
3 6 6.243.2 15.81.4 20.27.8 15.9.16 6.3.32 64
(A) 1
Solution :
3
(B) 2
(C) 3
The numerator is of the form a 3 b 3 3ab a b a b Where, a = 18 and b = 7 Denominator can be written as
(D) 4 3
Nr = (18 + 7)3 = (25)3 6
3 6 6 C 1 .3 5 .21 6 C 2 .3 4 .2 2 6 C 3 .3 3 .2 3 6 C 4 3 2 .2 4 6 C 5 3 .2 5 6 C 6 2 6 3 2 5 6 25
Nr (25) 3 1 Dr (25) 3
3
Ans.
1