हिंदी

Prove the statement by using the Principle of Mathematical Induction: For any natural number n, 7n – 2n is divisible by 5.

Advertisements
Advertisements

प्रश्न

Prove the statement by using the Principle of Mathematical Induction:

For any natural number n, 7n – 2n is divisible by 5.

प्रमेय
Advertisements

उत्तर

P(n) = 7n – 2n is divisible by 5.

So, substituting different values for n, we get,

P(0) = 70 – 20 = 0 Which is divisible by 5.

P(1) = 71 – 21 = 5 Which is divisible by 5.

P(2) = 72 – 22 = 45 Which is divisible by 5.

P(3) = 73 – 23 = 335 Which is divisible by 5.

Let P(k) = 7k – 2k be divisible by 5.

So, we get,

⇒ 7k – 2k = 5x

Now, we also get that,

⇒ P(k + 1)= 7k+1 – 2k+1

= (5 + 2)7k – 2(2k)

= 5(7k) + 2(7k – 2k)

= 5(7k) + 2(5x) Which is divisible by 5.

⇒ P(k + 1) is true when P(k) is true.

Therefore, by Mathematical Induction, P(n) = 7n – 2n is divisible by 5 is true for each natural number n.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Principle of Mathematical Induction - Exercise [पृष्ठ ७१]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
अध्याय 4 Principle of Mathematical Induction
Exercise | Q 7 | पृष्ठ ७१

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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 ∈ Nx2n – y2n is divisible by x y.


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


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


12 + 32 + 52 + ... + (2n − 1)2 = \[\frac{1}{3}n(4 n^2 - 1)\]

 

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


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

 

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

 

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


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

 

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

  

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

 


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

 

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


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:

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


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

(24n−1) is divisible by 15


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

8 + 17 + 26 + … + (9n – 1) = `"n"/2(9"n" + 7)`


Answer the following:

Given that tn+1 = 5tn − 8, t1 = 3, prove by method of induction that tn = 5n−1 + 2


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:

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


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.


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"):}`


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`


Give an example of a statement P(n) which is true for all n. Justify your answer. 


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:

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


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.


Prove the statement by using the Principle of Mathematical Induction:

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


A sequence d1, d2, d3 ... is defined by letting d1 = 2 and dk = `(d_(k - 1))/"k"` for all natural numbers, k ≥ 2. Show that dn = `2/(n!)` for all n ∈ N.


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.


Consider the statement: “P(n) : n2 – n + 41 is prime." Then which one of the following is true?


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×