JEE Advanced (Fill in the Blanks): Mathematical Induction & Binomial Theorem | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. The larger of 99⁵⁰ + 100⁵⁰ and 101⁵⁰ is ................ (1982 - 2 Marks) 

Ans.  (101)⁵⁰

Sol.  Consider (101)⁵⁰ – {(99)⁵⁰ + (100)⁵⁰} = (100 + 1)⁵⁰  –  (100 – 1)⁵⁰ – (100)⁵⁰
= (100)⁵⁰ [(1+ 0.0 1)⁵⁰ – (1– 0.01)⁵⁰ – 1]
= (100)⁵⁰ [2 (⁵⁰C₁(0.01) + ⁵⁰C₃(0.01)³  + ....) – 1]
= (100)⁵⁰ [2 (⁵⁰C₃(0.01)³  + ....)] >  0
∴ (101)⁵⁰ > (99)⁵⁰ + (100)⁵⁰ 
 ∴  (101)⁵⁰   is greater.

Q.2. The sum of the coefficients of the plynomial (1 + x – 3x²)²¹⁶³ is ................ (1982 - 2 Marks) 

Ans. -1

Sol.  If we put  x = 1 in the expansion of (1+ x – 3x2)²¹⁶³ = A₀ + A₁x + A₂x² + ...
we will get the sum of coefficients of given polynomial, which clearly comes to be – 1.

Q.3. If (1 + ax )ⁿ = 1 + 8x + 24x² + ....  then a = ...... and n = ............... (1983 - 2 Marks) 

Ans. a = 2, n = 4

Sol.  (1 + ax)ⁿ = 1 + 8x + 24x² + ...
⇒ (1 + ax)ⁿ = 1 + nxa +

= 1 + 8 x + 24x 2+ ...
Comparing like powers of  x we get

nax = 8x ⇒ na = 8 ....(1)

   ....(2)

Solving (1) and (2),  n = 4,  a = 2
 

Q.4. Let n be 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 .............. (1994 -  2 Marks) 

Ans. 7

 Sol.  We know that for a +ve integer n (1 + x)ⁿ = ⁿC₀ + ⁿC₁ x + ⁿC₂ x² + ......+ ⁿC_n xⁿ
ATQ coefficients of 2^nd, 3^nd, and 4^th terms are in A.P. i.e.ⁿC₁, ⁿC₂, ⁿC₃ are in A.P.
⇒ 2.ⁿC₂ =  ⁿC₁ +  ⁿC₃

⇒ n² – 9n + 14 = 0

⇒ (n – 7) (n – 2) = 0 ⇒  n = 7  or  2
But for the existance of 4^th term, n = 7.

Q.5. The sum of the rational terms in th e expansion of     is ................ (1997 - 2 Marks)

Ans. 41

Sol.  Let T_{r +1} be the general term in the expansion of

Let T_{r +1}will be rational if 2^5–r/2 and 3^r/5 are rational numbers.

⇒  are integers.

⇒ r = 0 and r = 10   ⇒  T₁ and T₁₁ are rational terms.
⇒ Sum of T₁  and T₁₁ = ¹⁰C₀25 – 0.30 + ¹⁰C₁₀2^5–5.3²
= 1.32.1 + 1.1.9 = 32 + 9 = 41