INDUCTION-QUIZ-6

MATHEMATICS INDUCTION QUIZ-6

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. If a₁=1 and a_n=na_n-1 for all positive integer n≥2, then a₅ is equal to

•  125
•  120
•  100
•  24

Solution

Q2.7²ⁿ+3^n-1.2^3n-3 is divisible by

•  24
•  25
•  9
•  13

Q3.  If n is a positive integer, then 5²ⁿ⁺²-24n-25 isdivisible by ∑y=50,∑xy=220,∑x² =200,∑y² =262,n=10 is

•  574
•  575
•  675
•  576

Solution

Q4. The product of three consecutive natural numbers is divisible by

•  5
•  7
•  6
•  4

Solution

Q5.If x^2n-1+y^2n-1 is divisible by x+y, if n is

•  A positive integer
•  An even positive integer
•  An odd positive integer
•  None of the above

Solution
 

Q6. If n is an even positive integer, then aⁿ+bⁿ is divisible by

•  a+b
•  a-b
•  a²-b²
•  None of these

Q7.The inequality n!>2^n-1 is true for

• n>2
•  n∈N
•  n>3
•  None of these

Solution

Q8.For all n∈N,3n⁵+5n³+7n is divisible by

•  3
•  5
•  10
•  15

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

Q9.The sum to n terms of the series 1³+3³+5³+⋯is

•  n²(n²-1)
•  n²(2n²-1)
•  n²(2n²+1)
•  n²(n²+1)

Solution
n²(2n²-1) gives the sum of the series for n=1,2, etc.

Q10. 10ⁿ+3(4ⁿ⁺² )+5 is divisible by (n∈N)

•  7
•  5
•  9
•  17

Solution