## Saturday, August 30, 2008

### Binomial theorem - July Dec Revision

If x and a are real numbers, then for all n Є N

(x+a)n
= nC0xna0 + nC1xn-1a1 +nC2xn-2a2 + ...+nCrxn-rar+ ...+nCn-1x1an-1+nCnx0an

(x+a)n
= (r = 0 to n)Σ nCrxn-rar

Some conclusions

1. Total number of terms in the expansion = n+1

2. The sum of indices of x and a in each term is n.

3. The coefficients of terms equidistant from the beginning and the end are equal.

4. (x-a)n
= (r = 0 to n)Σ ((-1) r*nCrxn-rar

The terms in the expansion of(x-a)n are alternatively positive and negative, the last term is positive or negative according as n is even or odd.

5. (1+x) n = (r = 0 to n)Σ nCrxr

you get it by putting x =1 and a = x in the expression for (x+a)n.

6. (x+1) n = (r = 0 to n)Σ nCrxn-r

7. (1-x) n = (r = 0 to n)Σ(-1)r* nCrxr

8. (x+a) n +(x-a) n = 2[nC0xna0 +
nC2xn-2a2 +
nC4xn-4a4+ ...]

9. General term in a binomial expansion

(r+1) term in (x+a) n
= nCrxn-rar

10. Another form of binomial expansion

(x+a)n
= (r = 0 to n) and r+s = n Σ (n!/r!s!) xras

11. Coefficient of (r+1)th term in the binomial expression of (1+x)n is nCr

12. Algorithm to find the greatest term

(i) Write term r+1 = T(r+1) and term r = T® from the given expression.
(ii) Find T(r+1)/T®
(iii) Put T(r+1)/T®>1
(iv) Solve the inequality for r to get an inequality of the form rm

If m is an integer, the mth and (m+1)th terms are equal in magnitude and these two are the greatest terms.

If m is not an integer, then obtain integral part of m, say, k. In this case (k+1) term is the greatest term

13 Properties binomial coefficients

(i) the sum of binomial coefficients in the expansion of (1+x)n is 2n

(ii) the sum of odd binomial coefficients in the expansion of (1+x)n is equal to the sum of the coefficients of even terms and each is equal to 2n-1.

14. Middle terms in binomial expression

If n is even the ((n/2) +1) th term is middle term.
If n is odd then ((n+1)/2) th and ((n+3)/2)th terms are two middle terms

Multinomial theorem

(x1+x2)n = (r1 = 0 to n) and r1+r2 = nΣ (n!/r1!r2!) x1r1x2r2

Past JEE Problem

The coefficient of x4 in [(x/2)- (3/x²)]10 is

a. 405/256
b. 504/259
c. 450/263
d. none of these

(JEE 1983)

Use the formula

(r+1) term in (x+a) n
= nCrxn-rar

to find r which will give x4 term in expansion.

For some more problems of JEE

http://iit-jee-maths-aps.blogspot.com/2008/08/binomial-theorem-model-problems-past.html