हिंदी

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

Advertisements
Advertisements

प्रश्न

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

योग
Advertisements

उत्तर

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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 12: Mathematical Induction - Exercise 12.2 [पृष्ठ २८]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 12 Mathematical Induction
Exercise 12.2 | Q 22 | पृष्ठ २८

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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

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: 32n + 2 – 8n– 9 is divisible by 8.


If P (n) is the statement "n2 − n + 41 is prime", prove that P (1), P (2) and P (3) are true. Prove also that P (41) is not true.


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

 

1.2 + 2.22 + 3.23 + ... + n.2= (n − 1) 2n+1+2

 

1.3 + 2.4 + 3.5 + ... + n. (n + 2) = \[\frac{1}{6}n(n + 1)(2n + 7)\]

 

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


Prove that 1 + 2 + 22 + ... + 2n = 2n+1 - 1 for all \[\in\] N .

 

\[\frac{1}{2}\tan\left( \frac{x}{2} \right) + \frac{1}{4}\tan\left( \frac{x}{4} \right) + . . . + \frac{1}{2^n}\tan\left( \frac{x}{2^n} \right) = \frac{1}{2^n}\cot\left( \frac{x}{2^n} \right) - \cot x\] for all n ∈ and  \[0 < x < \frac{\pi}{2}\]

 


\[1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + . . . + \frac{1}{n^2} < 2 - \frac{1}{n}\] for all n ≥ 2, n ∈ 

 


x2n−1 + y2n−1 is divisible by x + y for all n ∈ N.

 

\[\text{ Prove that }  \frac{1}{n + 1} + \frac{1}{n + 2} + . . . + \frac{1}{2n} > \frac{13}{24}, \text{ for all natural numbers } n > 1 .\]

 


\[\text{ Let } P\left( n \right) \text{ be the statement } : 2^n \geq 3n . \text{ If } P\left( r \right) \text{ is true, then show that } P\left( r + 1 \right) \text{ is true . Do you conclude that } P\left( n \right)\text{  is true for all n }  \in N?\]


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

2 + 4 + 6 + ..... + 2n = n (n+1)


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

13 + 33 + 53 + .... to n terms = n2(2n2 − 1)


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

5 + 52 + 53 + .... + 5n = `5/4(5^"n" - 1)`


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:

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


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

`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 loga xn = n logax, x > 0, n ∈ N


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

1 + 3 + 5 + ... + (2n – 1) = n2 


The distributive law from algebra says that for all real numbers c, a1 and a2, we have c(a1 + a2) = ca1 + ca2.

Use this law and mathematical induction to prove that, for all natural numbers, n ≥ 2, if c, a1, a2, ..., an are any real numbers, then c(a1 + a2 + ... + an) = ca1 + ca2 + ... + can.


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"):}`


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 ______.


Prove the statement by using the Principle of Mathematical Induction:

4n – 1 is divisible by 3, for each natural number n.


Prove the statement by using the Principle of Mathematical Induction:

n3 – 7n + 3 is divisible by 3, for all natural numbers n.


Prove the statement by using the Principle of Mathematical Induction:

n3 – n is divisible by 6, for each natural number n ≥ 2.


Prove the statement by using the Principle of Mathematical Induction:

n2 < 2n for all natural numbers n ≥ 5.


A sequence a1, a2, a3 ... is defined by letting a1 = 3 and ak = 7ak – 1 for all natural numbers k ≥ 2. Show that an = 3.7n–1 for all natural numbers.


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×