72n + 23n−3. 3n−1 Is Divisible by 25 for All N ∈ N.

Question

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

Solution

Let P(n) be the given statement.
Now,

\[P(n): 7^{2n} + 2^{3n - 3} . 3^{n - 1} \text{ is divisible by } 25 . \]
\[\text{ Step1} : \]
\[P(1): 7^2 + 2^{3 - 3} . 3^{1 - 1} = 49 + 1 = 50 \]
\[\text{ It is divisible by 25} . \]
\[\text{ Thus, P(1) is true} \]
\[\text{ Step2: Let } P\left( m \right) \text{ be true . } \]
\[\text{ Now } , \]
\[ 7^{2m} + 2^{3m - 3} . 3^{m - 1} \text{ is divisible by} 25 . \]
\[Suppose: \]
\[ 7^{2m} + 2^{3m - 3} . 3^{m - 1} = 25\lambda . . . (1)\]
\[\text{ We have to show that } P\left( m + 1 \right) \text{ is true whenever P(m) is true} . \]
\[\text{ Now, } \]
\[P\left( m + 1 \right) = 7^{2m + 2} + 2^{3m} . 3^m \]
\[ = 7^{2m + 2} + 7^2 . 2^{3m - 3} . 3^{m - 1} - 7^2 . 2^{3m - 3} . 3^{m - 1} + 2^{3m} . 3^m \]
\[ = 7^2 \left( 7^{2m} + 2^{3m - 3} . 3^{m - 1} \right) + 2^{3m} . 3^m \left( 1 - \frac{49}{24} \right)\]
\[ = 7^2 \times 25\lambda - 2^{3m} . 3^m \times \frac{25}{2^3 . 3^1} \left[ \text{ Using } (1) \right]\]
\[ = 25\left( 49\lambda - 2^{3m - 3} . 3^{m - 1} \right) \]
\[\text{ It is divisible by } 25 . \]
\[\text{ Thus, } P\left( m + 1 \right) \text{ is true .} \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n } \in N .\]

shaalaa.com

Principle of Mathematical Induction

Is there an error in this question or solution?

Q 25 Q 24 Q 26

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 25 | Page 28

Video TutorialsVIEW ALL [1]

view
Video Tutorials For All Subjects
Principle of Mathematical Induction

video tutorial00:17:42

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: 1.2.3 + 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: 1.2 + 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: `1/2 + 1/4 + 1/8 + ... + 1/2^n = 1 - 1/2^n`

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

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

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: n (n + 1) (n + 5) is a multiple of 3.

Prove the following by using the principle of mathematical induction for all n ∈ N: x²ⁿ – y²ⁿ is divisible by x + y.

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

Given an example of a statement P (n) such that it is true for all n ∈ N.

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

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

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.

11ⁿ⁺² + 12²ⁿ⁺¹ is divisible by 133 for all n ∈ N.

Prove that 1 + 2 + 2² + ... + 2ⁿ = 2ⁿ⁺¹ - 1 for all n \[\in\] 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 ∈ N and \[0 < x < \frac{\pi}{2}\]

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

Show by the Principle of Mathematical induction that the sum S_n 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 \[S_n = \binom{\frac{n \left( n + 1 \right)^2}{2}, \text{ if n is even} }{\frac{n^2 \left( n + 1 \right)}{2}, \text{ if n is odd } }\]

\[\text{ Using principle of mathematical induction, prove that } \sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + . . . + \frac{1}{\sqrt{n}} \text{ for all natural numbers } n \geq 2 .\]

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

3 + 7 + 11 + ..... + to n terms = n(2n+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:

`1/(1.3) + 1/(3.5) + 1/(5.7) + ... + 1/((2"n" - 1)(2"n" + 1)) = "n"/(2"n" + 1)`

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

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

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

2n + 1 < 2ⁿ, for all natual numbers n ≥ 3.

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

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

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`