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Test: Principle Of Mathematical Induction- 1 - JEE MCQ


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25 Questions MCQ Test Chapter-wise Tests for JEE Main & Advanced - Test: Principle Of Mathematical Induction- 1

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Test: Principle Of Mathematical Induction- 1 - Question 1

Let P(n) be a statement and let P(n)⇒P(n+1) for all natural numbers n, then P(n) is true for

Detailed Solution for Test: Principle Of Mathematical Induction- 1 - Question 1

Given, P(n) = P(n+1)∀n∈N
Substituting n−1 in place of n,
P(n−1) = P(n)
Thus if P(k) is true for some k ∈ N, then it is true for k−1 and k+1.
Thus, it is true ∀ k ∈ N

Test: Principle Of Mathematical Induction- 1 - Question 2

Let P(n) be a statement 2n<n!, where n is a natural number, then P(n) is true for

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Test: Principle Of Mathematical Induction- 1 - Question 3

If x > -1, then the statement (1+x)n>1+nx is true for

Test: Principle Of Mathematical Induction- 1 - Question 4

The smallest +ve integer n , for whichn!

Test: Principle Of Mathematical Induction- 1 - Question 5

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

Test: Principle Of Mathematical Induction- 1 - Question 6

x(xn-1 - nan-1)+ an (n-1) is divisible by (x-a)2 for

Test: Principle Of Mathematical Induction- 1 - Question 7

The greatest positive integer, which divides n (n + 1) (n + 2) (n + 3) for all n ∈ N, is

Test: Principle Of Mathematical Induction- 1 - Question 8

Let P (n) denote the statement n2+n is odd, It is seen that P(n) ⇒ P(n+1), therefore P (n) is true for all

Test: Principle Of Mathematical Induction- 1 - Question 9

The greatest positive integer, which divides (n+1)(n+2)(n+3)..................(n+r) ∀n∈W, is

Detailed Solution for Test: Principle Of Mathematical Induction- 1 - Question 9


P(n) = (n+1)(n+2).........(n+r)
 
Now find p(1)
P(1) = r! (r+1)
 
Similarly
P(k) will be divisible by r!
P(k+1) is also divisible by r!
 
Hence p(n) will also be divisible by r!.

Test: Principle Of Mathematical Induction- 1 - Question 10

The statement P (n) : “1 X 1! + 2 X 2! + 3 X 3! + …..+ n X n! = (n + 1) !..... 1 “ is

Test: Principle Of Mathematical Induction- 1 - Question 11

A student was asked to prove a statement P (n) by method of induction. He proved that P (3) is true and that P(n) ⇒ P(n+1) for all natural numbers n. On the basis of this he could conclude that P (n) is true

Test: Principle Of Mathematical Induction- 1 - Question 12

32n+2−8n−9is divisible by 64 for all

Detailed Solution for Test: Principle Of Mathematical Induction- 1 - Question 12

32n+2−8n−9=9n+1−9−8n32n+2−8n−9=9n+1−9−8n

=9(9n−1)−8n=9(9n−1)−8n

=9((8+1)n−1)−8n=9((8+1)n−1)−8n

=9∑k=1n(nk)8k−8n=9∑k=1n(nk)8k−8n

==9∑k=2n(nk)8k+64n9∑k=2n(nk)8k+64n

is a multiple of 64

Test: Principle Of Mathematical Induction- 1 - Question 13

The inequality n!>2n−1 is true

Test: Principle Of Mathematical Induction- 1 - Question 14

The statement 3n>4n is true for all

Test: Principle Of Mathematical Induction- 1 - Question 15

The statement 2n>3n is true for all

Test: Principle Of Mathematical Induction- 1 - Question 16

The statement 2n+2<3n is true for all

Test: Principle Of Mathematical Induction- 1 - Question 17

The smallest positive integer for which The statement 3n+1<4n is true for

Test: Principle Of Mathematical Induction- 1 - Question 18

For all n ∈ N , 49n+16n−1 is divisible by

Test: Principle Of Mathematical Induction- 1 - Question 19

The digit in the unit’s place of the number 183! + 3183 is

Detailed Solution for Test: Principle Of Mathematical Induction- 1 - Question 19

Test: Principle Of Mathematical Induction- 1 - Question 20

72n+3n−1.23n−3 is divisible by

Test: Principle Of Mathematical Induction- 1 - Question 21

If n ∈ N then n3+2n is divisible by

Detailed Solution for Test: Principle Of Mathematical Induction- 1 - Question 21

Induction works in the following way: If you show that the result being true for any integer implies it is true for the next, then you need only show that it is true for n=1 for it to be true for n=2 and then n=3 and so on.

Step 1: Show true for n=1

For n=1, n^3+2n=(1)^3+2(1)

n^3+2n=3

3 is definitely divisible by 3 so the statement is true for n=1.

Step 2: Assume true for n=k

We assume that for any integer k, n^3+2n is divisible by 3. We can write this mathematically as:

k^3+2k=3m, where m is an integer

Step 3: Show true for k+1

For n=k+1,

n^3+2n=(k+1)^3+2(k+1)

=(k^3+3k^2+3k+1)+2k+2

=(k^3+2k)+3(k^2+k+1)

Subbing in from part 2 for (k^3+2k), we get:

n^3+2n=3m+3(k^2+k+1)

=3(m+k^2+k+1)

which is divisible by 3.

 

This means that the statement being true for n=k implies the statement is true for n=k+1, and as we have shown it to be true for n=1 the proof of the statement follows by induction.

Test: Principle Of Mathematical Induction- 1 - Question 22

If n is a positive integer , then 2.7n+3.5n−5 is divisible by

Test: Principle Of Mathematical Induction- 1 - Question 23

If n is a +ve integer, then 4n−3n−5 is divisible by

Detailed Solution for Test: Principle Of Mathematical Induction- 1 - Question 23

Test: Principle Of Mathematical Induction- 1 - Question 24

If n is a +ve integer, then 2.42n+1+33n+1 is divisible by

Test: Principle Of Mathematical Induction- 1 - Question 25

If n is a +ve integer, then 10n +3.4n+2 + 5 is divisible by

Detailed Solution for Test: Principle Of Mathematical Induction- 1 - Question 25

Sol : Let P(n) = 10n + 3.4n+2 + 5 is divisible by 9.

Step 1 : for n =1  we have
P(1) ; 10 + 3x64 + 5 = 207 = 9x23
Which is divisible by 9 .
∴ P(1) is true .

Step 2 :For n =k assume that P(k) is true .
Then P(k) : 10k + 3.4k+2 + 5 is divisible by 9.
   10k + 3.4k+2 + 5  = 9m
10k = 9m - 3.4k+2 - 5  ----------(1)

Step 3 ;
We have to to prove that P(k+1) is divisible by 9 for n = k + 1.

P(k + 1) : 10k+1 + 3.4k+1+2 + 5
            = 10 x 10k + 3.4k+2 .4+ 5 
            = 10 ( 9m - 3.4k+2 - 5 ) + 3.4k+2 .4+ 5 
            = 90m - 30 4k+2 - 50 + 12.4k+2 + 5    
            = 90m - 18 4k+2 - 45
            = 9( 10m - 2.4k+2 - 5 )
which is divisible by 9 .
∴ P (k +1 ) is true .
Hence by the Principle of mathematical induction P(n) is true for all n∈N.

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