Binomial theorem - July Dec Revision

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

(x+a)ⁿ
= ⁿC₀xⁿa⁰ + ⁿC₁x^n-1a¹ +ⁿC₂x^n-2a² + ...+ⁿC_rx^n-ra^r+ ...+ⁿC_n-1x¹a^n-1+ⁿC_nx⁰aⁿ

(x+a)ⁿ
= (r = 0 to n)Σ ⁿC_rx^n-ra^r


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_*ⁿC_rx^n-ra^r

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 ⁿC_rx^r

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 ⁿC_rx^n-r

7. (1-x) ⁿ = (r = 0 to n)Σ(-1)^r_* ⁿC_rx^r

8. (x+a) ⁿ +(x-a) ⁿ = 2[ⁿC₀xⁿa⁰ +
ⁿC₂x^n-2a² +
ⁿC₄x^n-4a⁴+ ...]

9. General term in a binomial expansion

(r+1) term in (x+a) ⁿ
= ⁿC_rx^n-ra^r

10. Another form of binomial expansion

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

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

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 2^n-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!) x1^r1x2^r2

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)

Answer: a

Use the formula

(r+1) term in (x+a) ⁿ
= ⁿC_rx^n-ra^r

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