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. - Mathematics

Question

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.

Theorem

Solution

P(n) = xⁿ – yⁿ is divisible by x – y, x integers with x ≠ y.

So, substituting different values for n, we get,

P(0) = x⁰ – y⁰ = 0 Which is divisible by x − y.

P(1) = x − y Which is divisible by x − y.

P(2) = x² – y²

= (x + y)(x − y) Which is divisible by x − y.

P(3) = x³ – y³

= (x − y)(x² + xy + y²) Which is divisible by x − y.

Let P(k) = x^k – y^k be divisible by x – y;

So, we get,

⇒ x^k – y^k = a(x − y).

Now, we also get that,

⇒ P(k + 1) = x^k+1 – y^k+1

= x^k(x − y) + y(x^k − y^k)

= x^k(x − y) + y^a(x − y) Which is divisible by x − y.

⇒ P(k + 1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) xⁿ – yⁿ is divisible by x – y, where x integers with x ≠ y which is true for any natural number n.

shaalaa.com

Principle of Mathematical Induction

Is there an error in this question or solution?

Q 8 Q 7 Q 9

Chapter 4: Principle of Mathematical Induction - Exercise [Page 71]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 11

Chapter 4 Principle of Mathematical Induction
Exercise | Q 8 | Page 71

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+ 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 + 2.3 + 3.4+ ... + n(n+1) = `[(n(n+1)(n+2))/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`

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

Give an example of a statement P(n) which is true for all n ≥ 4 but P(1), P(2) and P(3) are not true. Justify your answer.

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

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

\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]

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

\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . + \frac{1}{2^n} = 1 - \frac{1}{2^n}\]

a + ar + ar² + ... + arⁿ⁻¹ = \[a\left( \frac{r^n - 1}{r - 1} \right), r \neq 1\]

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.

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

Given \[a_1 = \frac{1}{2}\left( a_0 + \frac{A}{a_0} \right), a_2 = \frac{1}{2}\left( a_1 + \frac{A}{a_1} \right) \text{ and } a_{n + 1} = \frac{1}{2}\left( a_n + \frac{A}{a_n} \right)\] for n ≥ 2, where a > 0, A > 0.
Prove that \[\frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1}\]

7 + 77 + 777 + ... + 777 \[{. . . . . . . . . . .}_{n - \text{ digits } } 7 = \frac{7}{81}( {10}^{n + 1} - 9n - 10)\]

\[\frac{n^{11}}{11} + \frac{n^5}{5} + \frac{n^3}{3} + \frac{62}{165}n\] is a positive integer for all n ∈ N.

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

Prove that the number of subsets of a set containing n distinct elements is 2ⁿ, for all n \[\in\] N .

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