English

52n+2 −24n −25 is Divisible by 576 for All N ∈ N.

Advertisements
Advertisements

Question

52n+2 −24n −25 is divisible by 576 for all n ∈ N.

 
Advertisements

Solution

Let P(n) be the given statement.
Now,

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

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

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 12 Mathematical Induction
Exercise 12.2 | Q 21 | Page 28

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

`1 + 3 + 3^2 + ... + 3^(n – 1) =((3^n -1))/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 + 2.22 + 3.22 + … + n.2n = (n – 1) 2n+1 + 2


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

`a + ar + ar^2 + ... + ar^(n -1) = (a(r^n - 1))/(r -1)`

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 ∈ Nn (n + 1) (n + 5) is a multiple of 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 ∈ Nx2n – y2n is divisible by x y.


Prove the following by using the principle of mathematical induction for all n ∈ N: 41n – 14n is a multiple of 27.


If P (n) is the statement "2n ≥ 3n" and if P (r) is true, prove that P (r + 1) is true.

 

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.

 

\[\frac{1}{3 . 5} + \frac{1}{5 . 7} + \frac{1}{7 . 9} + . . . + \frac{1}{(2n + 1)(2n + 3)} = \frac{n}{3(2n + 3)}\]


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


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

 

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

 

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

 


52n −1 is divisible by 24 for all n ∈ N.


32n+7 is divisible by 8 for all n ∈ N.

 

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


2.7n + 3.5n − 5 is divisible by 24 for all n ∈ N.


\[\frac{n^7}{7} + \frac{n^5}{5} + \frac{n^3}{3} + \frac{n^2}{2} - \frac{37}{210}n\] is a positive integer for all n ∈ N.  

 


Show by the Principle of Mathematical induction that the sum Sn of then terms of the series  \[1^2 + 2 \times 2^2 + 3^2 + 2 \times 4^2 + 5^2 + 2 \times 6^2 + 7^2 + . . .\] is given by \[S_n = \binom{\frac{n \left( n + 1 \right)^2}{2}, \text{ if n is even} }{\frac{n^2 \left( n + 1 \right)}{2}, \text{ if n is odd } }\]

 


\[\text{ A sequence }  a_1 , a_2 , a_3 , . . . \text{ is defined by letting }  a_1 = 3 \text{ and } a_k = 7 a_{k - 1} \text{ for all natural numbers } k \geq 2 . \text{ Show that } a_n = 3 \cdot 7^{n - 1} \text{ for all } n \in N .\]


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

3n − 2n − 1 is divisible by 4


Answer the following:

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

`1/(3.4.5) + 2/(4.5.6) + 3/(5.6.7) + ... + "n"/(("n" + 2)("n" + 3)("n" + 4)) = ("n"("n" + 1))/(6("n" + 3)("n" + 4))`


Answer the following:

Prove by method of induction 152n–1 + 1 is divisible by 16, for all n ∈ N.


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. 


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


Prove the statement by using the Principle of Mathematical Induction:

23n – 1 is divisible by 7, for all natural numbers n.


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:

2 + 4 + 6 + ... + 2n = n2 + n for all natural numbers n.


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


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


If xn – 1 is divisible by x – k, then the least positive integral value of k is ______.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×