INDUCTION-QUIZ-7

MATHEMATICS INDUCTION QUIZ-7

Dear Readers,

As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background. 

Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. 

Q1. For all positive integral values of n,3²ⁿ-2n+1 is divisible by

•  2
•  4
•  8
•  12

Solution

Q2.The sum of the cubes of three consecutive natural numbers is divisible by

•  7
•  9
•  25
•  26

Q3.  For all n∈N,∑n ∑y=50,∑xy=220,∑x² =200,∑y² =262,n=10 is

•  
• 
  
  
•  
• 
  
  
•  
• 
  
  
•  None of these

Q4. If P(n):2+4+6…+(2n),n∈N, then P(k)=k(k+1)+2implies P(k)=(k+1)(k+2)+2 is true for all k∈N. So, statement P(n)=n(n+1)+2 is true for

•  n≥1
•  n≥2
•  n≥3
•  None of these

Solution

Q5.The nth term of the series 4+14+30+52+80+114+⋯ is

•  5n-1
•  2n²+2n
•  3n²+n
•  2n²+2

Solution
  We observe that 3n²+n gives various terms of the series by putting n=1,2,3,…

Q6. 

•  a(100)≤100
•  a(100)>100
•  a(200)≤100
•  None of these

Solution

Q7.If P(n)=2+4+6+...+2n,n∈N,then P(k)=k(k+1)+2⇒P(k+1)=(k+1)(k+2)+2 for all k∈N. So, we can conclude that P(n)=n(n+1)+2 for

• All n∈N
•  n>1
•  n>2
•  Nothing can be said

Solution
It is obvious, nothing can be said.

Q8.The number aⁿ-bⁿ (a,b are distinct rational numbers and n∈N) is always divisible by

•  a-b
•  a+b
•  2a-b
•  a-2b

Solution
Putting n=1,2,3…, it can be checked that 3n⁵+5n³+7n is divisible by 15

Q9.If P(n) is a statement such that P(3) is true. Assuming P(k) is true ⇒P(k+1) is true for all k≥3, then P(n) is true

•  For all n
•  For n≥3
•  For n>4
•  n²(n²+1)

Solution
Since, P(3) is true. Assume P(k) is true ⇒P(k+1) is true means, if P(3) is true ⇒P(4) is true ⇒P(5) is true and so on. So, statement is true for all n≥3.

Q10. If n∈N, then 3²ⁿ+7 is divisible by

•  3
•  8
•  9
•  11

Solution