English

If 10n + 3.4n+2 + k is divisible by 9 for all n ∈ N, then the least positive integral value of k is ______.

Advertisements
Advertisements

Question

If 10n + 3.4n+2 + k is divisible by 9 for all n ∈ N, then the least positive integral value of k is ______.

Options

  • 5

  • 3

  • 7

  • 1

MCQ
Fill in the Blanks
Advertisements

Solution

If 10n + 3.4n+2 + k is divisible by 9 for all n ∈ N, then the least positive integral value of k is 5.

Explanation:

Let P(n) = 10n + 3.4n + 2 + k is divisible by 9, ∀ n ∈  N

P(1) = 101 + 3.41 + 2 + k = 10 + 3.64 + k

= 10 + 192 + k = 202 + k must be divisible by 9.

If (202 + k) is divisible by 9 then k must be equal to 5.

202 + 5 = 207 which is divisible by 9.

= `207/9`

= 23

So, the least positive integral value of k = 5.

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Principle of Mathematical Induction - Exercise [Page 72]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 11
Chapter 4 Principle of Mathematical Induction
Exercise | Q 26 | Page 72

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Prove the following by using the principle of mathematical induction for all n ∈ N: 

`1+ 1/((1+2)) + 1/((1+2+3)) +...+ 1/((1+2+3+...n)) = (2n)/(n +1)`

Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2.3 + 2.3.4 + … + n(n + 1) (n + 2)  = `(n(n+1)(n+2)(n+3))/(4(n+3))`


Prove the following by using the principle of mathematical induction for all n ∈ N: 

1.3 + 2.3^3 + 3.3^3  +...+ n.3^n = `((2n -1)3^(n+1) + 3)/4`

Prove the following by using the principle of mathematical induction for all n ∈ N: 

1.3 + 3.5 + 5.7 + ...+(2n -1)(2n + 1) = `(n(4n^2 + 6n -1))/3`

Prove the following by using the principle of mathematical induction for all n ∈ N: 

`1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/((2n + 1)(2n +3)) = n/(3(2n +3))`

Prove the following by using the principle of mathematical induction for all n ∈ N: `1+2+ 3+...+n<1/8(2n +1)^2`


Prove the following by using the principle of mathematical induction for all n ∈ N: 102n – 1 + 1 is divisible by 11


If P (n) is the statement "n(n + 1) is even", then what is P(3)?


12 + 22 + 32 + ... + n2 =\[\frac{n(n + 1)(2n + 1)}{6}\] .

 

1 + 3 + 32 + ... + 3n−1 = \[\frac{3^n - 1}{2}\]

 

1 + 3 + 5 + ... + (2n − 1) = n2 i.e., the sum of first n odd natural numbers is n2.

 

1.2 + 2.3 + 3.4 + ... + n (n + 1) = \[\frac{n(n + 1)(n + 2)}{3}\]

 

32n+2 −8n − 9 is divisible by 8 for all n ∈ N.


n(n + 1) (n + 5) is a multiple of 3 for all n ∈ N.

 

72n + 23n−3. 3n−1 is divisible by 25 for all n ∈ N.

 

Prove that n3 - 7n + 3 is divisible by 3 for all n \[\in\] N .

  

\[\sin x + \sin 3x + . . . + \sin (2n - 1)x = \frac{\sin^2 nx}{\sin x}\]

 


\[\text{ A sequence } x_0 , x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_0 = 5 and x_k = 4 + x_{k - 1}\text{  for all natural number k . } \]
\[\text{ Show that } x_n = 5 + 4n \text{ for all n }  \in N \text{ using mathematical induction .} \]


Prove by method of induction, for all n ∈ N:

12 + 32 + 52 + .... + (2n − 1)2 = `"n"/3 (2"n" − 1)(2"n" + 1)`


Prove by method of induction, for all n ∈ N:

1.2 + 2.3 + 3.4 + ..... + n(n + 1) = `"n"/3 ("n" + 1)("n" + 2)`


Prove by method of induction, for all n ∈ N:

(23n − 1) is divisible by 7


Prove by method of induction, for all n ∈ N:

Given that tn+1 = 5tn + 4, t1 = 4, prove that tn = 5n − 1


Answer the following:

Prove, by method of induction, for all n ∈ N

12 + 42 + 72 + ... + (3n − 2)2 = `"n"/2 (6"n"^2 - 3"n" - 1)`


Answer the following:

Prove by method of induction loga xn = n logax, x > 0, n ∈ N


Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:

`sum_(t = 1)^(n - 1) t(t + 1) = (n(n - 1)(n + 1))/3`, for all natural numbers n ≥ 2.


Prove statement by using the Principle of Mathematical Induction for all n ∈ N, that:

`(1 - 1/2^2).(1 - 1/3^2)...(1 - 1/n^2) = (n + 1)/(2n)`, for all natural numbers, n ≥ 2. 


Define the sequence a1, a2, a3 ... as follows:
a1 = 2, an = 5 an–1, for all natural numbers n ≥ 2.

Use the Principle of Mathematical Induction to show that the terms of the sequence satisfy the formula an = 2.5n–1 for all natural numbers.


Let P(n): “2n < (1 × 2 × 3 × ... × n)”. Then the smallest positive integer for which P(n) is true is ______.


A student was asked to prove a statement P(n) by induction. He proved that P(k + 1) is true whenever P(k) is true for all k > 5 ∈ N and also that P(5) is true. On the basis of this he could conclude that P(n) is true ______.


State whether the following proof (by mathematical induction) is true or false for the statement.

P(n): 12 + 22 + ... + n2 = `(n(n + 1) (2n + 1))/6`

Proof By the Principle of Mathematical induction, P(n) is true for n = 1,

12 = 1 = `(1(1 + 1)(2*1 + 1))/6`. Again for some k ≥ 1, k2 = `(k(k + 1)(2k + 1))/6`. Now we prove that

(k + 1)2 = `((k + 1)((k + 1) + 1)(2(k + 1) + 1))/6`


Give an example of a statement P(n) which is true for all n. Justify your answer. 


Prove the statement by using the Principle of Mathematical Induction:

32n – 1 is divisible by 8, for all natural numbers n.


Prove the statement by using the Principle of Mathematical Induction:

For any natural number n, xn – yn is divisible by x – y, where x and y are any integers with x ≠ y.


Prove the statement by using the Principle of Mathematical Induction:

`sqrt(n) < 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(n)`, for all natural numbers n ≥ 2.


Prove the statement by using the Principle of Mathematical Induction:

1 + 5 + 9 + ... + (4n – 3) = n(2n – 1) for all natural numbers n.


Prove that, cosθ cos2θ cos22θ ... cos2n–1θ = `(sin 2^n theta)/(2^n sin theta)`, for all n ∈ N.


Prove that number of subsets of a set containing n distinct elements is 2n, for all n ∈ N.


For all n ∈ N, 3.52n+1 + 23n+1 is divisible by ______.


If P(n): 2n < n!, n ∈ N, then P(n) is true for all n ≥ ______.


By using principle of mathematical induction for every natural number, (ab)n = ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×