Binomial Theorem - Model Problems - Past JEE

1. 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)

2. Let n be a positive integer. If the coefficient of 2nd, 3rd, and 4th terms in the expansion of (1 + x)ⁿ are in A.P. then the value of n is ________________ .
(JEE 1994)


3. If the binomial expansion of (a-b)ⁿ, n≥5, the sum of the 5th and 6th terms is zero. The a/b equals

a. (n-5)/6
b. (n-4)/5
c. 5/(n-4)
d. 6/(n-5)

(JEE 2001)


4. The coefficient of t²⁴ in (1+t²)¹²(1+t¹²)(1+²⁴) is

a. ¹²C₆ + 3
b. ¹²C₆ + 1
c. ¹²C₆
d. ¹²C₆ + 2

(JEE 2003)


Answers and Solutions

1. 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)

Solution:
(r+1) th term of the expansion is given by
¹⁰C_r (x/2) ^10-r(-3/x²) ^r

= ¹⁰C_r (x) ^10-r(1/ x²)^r(1/2) ^10-r(-3) ^r

= ¹⁰C_r(-3) ^r(1/2) ^10-r(x) ^10-3r

In this term (x) ^10-3r is equal to x⁴
 10-3r = 4 or r = 2

So coefficient =
¹⁰C_r(-3) ^r(1/2) <sup>10-r

10</sup>C₂(-3) ²(1/2) ^10-2

= [10*9/2]*9*(1/2⁸)

= (45*9)/256
= 405/256




2. Let n be a positive integer. If the coefficients of 2nd, 3rd, and 4th terms in the expansion of (1 + x)ⁿ are in A.P. then the value of n is ________________ .
(JEE 1994)


Answer: 7

Solution:

The question requires concepts from binomial theorem and concept from arithmetic progression.

Coefficients of 2nd, 3rd, and 4th terms in the expansion of (1 + x)ⁿ are
ⁿC₁, ⁿC₂, ⁿC₃

According to arithmetic progression properties, three terms can be taken as a-d, a, and a+d. Hence sum of first and third terms = 2*second term

=> 2(ⁿC₂) = ⁿC₁ + ⁿC₃

=> 2 n(n-1)/2 = n + [n(n-1)(n-2)/6]

=> n-1 = 1 +[(n² – 3n+2)/6]

=> n² – 3n+2 = 6n-12
=> n² – 9n+14 = 0
=> (n-2)(n-7) = 0

Since 4th term will be there only when n>2, n is equal to 7.





3. If the binomial expansion of (a-b)ⁿ, n≥5, the sum of the 5th and 6th terms is zero. The a/b equals

a. (n-5)/6
b. (n-4)/5
c. 5/(n-4)
d. 6/(n-5)

(JEE 2001)

Answer: (b)

Solution

5th term = 4+1th term = ⁿC₄ a ^n-4(-b) ⁴

6th term = 5+1th term = ⁿC₅ a ^n-5(-b) ⁵

As sum of 5th terms and 6th terms is zero

ⁿC₄ a ^n-4(-b) ⁴ + ⁿC₅ a ^n-5(-b) ⁵ = 0

ⁿC₄ a ^n-4(b) ⁴ = ⁿC₅ a ^n-5(b) ⁵

a ^n-4(b) ⁴/ a ^n-5(b) ⁵ = ⁿC₅/ⁿC₄

a/b = [n!/(n-5)!(5!)]/[n!/(n-4)!(4!)

= (n-4)!/(5(n-5)!)
= (n-4)/5



4. The coefficient of t²⁴ in (1+t²)¹²(1+t¹²)(1+²⁴) is

a. ¹²C₆ + 3
b. ¹²C₆ + 1
c. ¹²C₆
d. ¹²C₆ + 2

(JEE 2003)

Answer: (d)

Solution:
Multiplying the last two expressions we have to find coefficient of t²⁴ in (1+t²)¹²(1+t¹²+²⁴)

t²⁴ term will come as a sum of three terms in the expanded expression

1st term is t²⁴ term in 1st expression multiplied by 1 in the second expression.
2nd term is t¹² term in 1st expression multiplied by t¹² term in the second expression.
3rd term is 1 in the first expression multiplied by t²⁴ term in the second expression.

Coefficients of 1 and 3 terms are 1 and 1.

Coefficient of 2 term = coefficient t¹² term in 1 st expression which is (1+t²)¹²

7th term or 6+1th term in the expansion will have t¹² term

Hence coefficient of 7th term = ¹²C₆

Therefore coefficient of t²⁴ term will be

¹²C₆ + 2