##
**
****MATHEMATICS INDUCTION QUIZ-2**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.

**MATHEMATICS INDUCTION QUIZ-2**

**Q1.**3+13+29+51+79+⋯ to n terms =

Solution

Clearly, n

Clearly, n

^{3}+2n^{2}gives the sum of the series for n=1,2,3 etc.**Q3.**The nth terms of the series 3+7+13+21+⋯ is ∑y=50,∑xy=220,∑x

^{2}=200,∑y

^{2}=262,n=10 is

Solution

Putting n=1,2,3…, we observe that 4n-1 is the nth term

Putting n=1,2,3…, we observe that 4n-1 is the nth term

**Q4.**10

^{n}+3(4

^{n+2})+5 is divisible by (n∈N)

Solution

For n=1,10

For n=1,10

^{n}+3∙4^{n+2}+5 =10+3∙4^{3}+5=207which is divisible by 9. ∴ By induction, the result is divisible by 9.**Q5.**If n∈N , then 11

^{n+2}+12

^{2n+1}is divisible by

On putting n=1 in 11

^{n+2}+12

^{2n+1}, we get 11

^{1+2}+12

^{2×1+1}=11

^{3}+12

^{3}=3059 Which is divisible by 133

**Q7.**Matrix A is such that A

^{2}=2A-I where I is the identity matrix, then for n≥2,A

^{n}is equal to

Solution

As we have A

As we have A

^{2}=2A-I ⟹ A

^{2}A=(2A-I)A=2A^{2}-IA ⟹ A

^{3}=2(2A-I)-IA=3A-2I Similarly, A

^{4}=4A-3I A

^{5}=5A-AI A

^{n}=nA-(n-1)I**Q8.**If n∈N, then n(n

^{2}-1) is divisible by

Solution

We have, n(n

We have, n(n

^{2}-1)=(n-1)(n+1), which is product of three consecutive natural numbers and hence divisible by 6**Q10.**Let S(k)=1+3+5….+(2k-1)=3+k

^{2}. Then, which of the following is true?

Solution