English

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

Advertisements
Advertisements

Question

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

Sum
Advertisements

Solution

Let P(n) be the given statement.
Now,

\[P(n): 5^{2n + 2} - 24n - 25 \text{ is divisible by 576 for all }  n \in N . \]
\[\text{ Step } 1: \]
\[P(1) = 5^{2 + 2} - 24 - 25 = 625 - 49 = 576 \]
\[\text{ It is divisible by }  576 . \]
\[\text{ Thus, P(1) is true}  . \]
\[\text{ Step2:}  \]
\[\text{ Let P(m) be true . } \]
\[Then, \]
\[ 5^{2m + 2} - 24m - 25 \text{ is divisible by } 576 . \]
\[\text { Let } 5^{2m + 2} - 24m - 25 = 576\lambda, \text{ where } \lambda \in N . \]
\[\text { We need to show that P(m + 1) is true whenever P(m) is true }  . \]
\[ \text{ Now, } \]
\[P(m + 1) = 5^{2m + 4} - 24(m + 1) - 25\]
\[ = 5^2 \times (576\lambda + 24m + 25) - 24m - 49\]
\[ = 25 \times 576\lambda + 600m + 625 - 24m - 49\]
\[ = 25 \times 576\lambda + 576m + 576\]
\[ = 576(25\lambda + m + 1) \]
\[\text{ It is divisible by } 576 . \]
\[\text{ Thus, P(m + 1) is true }  . \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n }  \in N . \] 

shaalaa.com
  Is there an error in this question or solution?
Chapter 12: Mathematical Induction - Exercise 12.2 [Page 28]

APPEARS IN

RD Sharma Mathematics [English] Class 11
Chapter 12 Mathematical Induction
Exercise 12.2 | Q 22 | Page 28

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

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


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+3/1)(1+ 5/4)(1+7/9)...`(1 + ((2n + 1))/n^2) = (n + 1)^2`

 

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

`1^2 + 3^2 + 5^2 + ... + (2n -1)^2 = (n(2n - 1) (2n + 1))/3`

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


Prove the following by using the principle of mathematical induction for all n ∈ N (2+7) < (n + 3)2


1 + 2 + 3 + ... + n =  \[\frac{n(n + 1)}{2}\] i.e. the sum of the first n natural numbers is \[\frac{n(n + 1)}{2}\] .


\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]

 


\[\frac{1}{3 . 7} + \frac{1}{7 . 11} + \frac{1}{11 . 5} + . . . + \frac{1}{(4n - 1)(4n + 3)} = \frac{n}{3(4n + 3)}\] 


1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) =\[\frac{n(4 n^2 + 6n - 1)}{3}\]

 

a + (a + d) + (a + 2d) + ... (a + (n − 1) d) = \[\frac{n}{2}\left[ 2a + (n - 1)d \right]\]

 


(ab)n = anbn for all n ∈ N. 

 

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

  

Let P(n) be the statement : 2n ≥ 3n. If P(r) is true, show that P(r + 1) is true. Do you conclude that P(n) is true for all n ∈ N


\[\frac{(2n)!}{2^{2n} (n! )^2} \leq \frac{1}{\sqrt{3n + 1}}\]  for all n ∈ N .


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


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

12 + 22 + 32 + .... + n2 = `("n"("n" + 1)(2"n" + 1))/6`


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:

3n − 2n − 1 is divisible by 4


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

(cos θ + i sin θ)n = cos (nθ) + i sin (nθ)


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

`[(1, 2),(0, 1)]^"n" = [(1, 2"n"),(0, 1)]` ∀ n ∈ N


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

`[(3, -4),(1, -1)]^"n" = [(2"n" + 1, -4"n"),("n", -2"n" + 1)], ∀  "n" ∈ "N"`


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

22n – 1 is divisible by 3.


Prove by the Principle of Mathematical Induction that 1 × 1! + 2 × 2! + 3 × 3! + ... + n × n! = (n + 1)! – 1 for all natural numbers n.


Show by the Principle of Mathematical Induction that the sum Sn of the n term of the series 12 + 2 × 22 + 32 + 2 × 42 + 52 + 2 × 62 ... is given by

Sn = `{{:((n(n + 1)^2)/2",",  "if n is even"),((n^2(n + 1))/2",",  "if n is odd"):}`


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


Prove the statement by using the Principle of Mathematical Induction:

2n < (n + 2)! for all natural number n.


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 that `1/(n + 1) + 1/(n + 2) + ... + 1/(2n) > 13/24`, for all natural numbers n > 1.


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


State whether the following statement is true or false. Justify.

Let P(n) be a statement and let P(k) ⇒ P(k + 1), for some natural number k, then P(n) is true for all n ∈ N.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×