मराठी

A Sequence X 0 , X 1 , X 2 , X 3 , . . . is Defined by Letting X 0 = 5 a N D X K = 4 + X K − 1 for All Natural Number K . Show that X N = 5 + 4 N for All N ∈ N Using Mathematical Induction . - Mathematics

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

\[\text{ Given: A sequence }  x_0 , x_1 , x_2 , x_3 , . . . \text{ is defined by letting }  x_0 = 5 \text{ and } x_k = 4 + x_{k - 1} \text{ for all natural number } k . \]

\[\text{ Let } P\left( n \right): x_n = 5 + 4n \text{ for all } n \in N . \]

\[\text{ Step I: For }  n = 0, \]

\[P\left( 0 \right): x_0 = 5 + 4 \times 0 = 5\]

\[\text{ So, it is true for }  n = 0 . \]

\[\text{ Step II: For } n = k, \]

\[\text{ Let } P\left( k \right): x_k = 5 + 4 \text{ k be true for some k } \in N . \]

\[\text{ Step III: For } n = k + 1, \]

\[P\left( k + 1 \right): x_{k + 1} = 4 + x_{k + 1 - 1} \]

\[ = 4 + x_k \]

\[ = 4 + 5 + 4k\]

\[ = 5 + 4\left( k + 1 \right)\]

\[\text{ So, it is also true for n }  = k + 1 . \]

\[\text{ Hence } , x_n = 5 + 4\text{ n for all  } n \in N .\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 12: Mathematical Induction - Exercise 12.2 [पृष्ठ २९]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 12 Mathematical Induction
Exercise 12.2 | Q 48 | पृष्ठ २९

व्हिडिओ 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

1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ...+ `1/(n(n+1)(n+2)) = (n(n+3))/(4(n+1) (n+2))`

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 ∈ 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 ∈ N

`1/1.4 + 1/4.7 + 1/7.10 + ... + 1/((3n - 2)(3n + 1)) = n/((3n + 1))`


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 "n(n + 1) is even", then what is P(3)?


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

 

If P (n) is the statement "n2 + n is even", and if P (r) is true, then 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 \[\frac{n(n + 1)}{2}\] .


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


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

 

(ab)n = anbn for all n ∈ N. 

 

72n + 23n−3. 3n−1 is divisible by 25 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 \[\frac{a_n - \sqrt{A}}{a_n + \sqrt{A}} = \left( \frac{a_1 - \sqrt{A}}{a_1 + \sqrt{A}} \right) 2^{n - 1}\]

 

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

 

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


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 \[S_n = \binom{\frac{n \left( n + 1 \right)^2}{2}, \text{ if n is even} }{\frac{n^2 \left( n + 1 \right)}{2}, \text{ if n is odd } }\]

 


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

 


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

12 + 22 + 32 + .... + n2 = `("n"("n" + 1)(2"n" + 1))/6`


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

12 + 32 + 52 + .... + (2n − 1)2 = `"n"/3 (2"n" − 1)(2"n" + 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:

`1/(3.5) + 1/(5.7) + 1/(7.9) + ...` to n terms = `"n"/(3(2"n" + 3))`


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:

`sum_(t = 1)^(n - 1) t(t + 1) = (n(n - 1)(n + 1))/3`, 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.


Let P(n): “2n < (1 × 2 × 3 × ... × n)”. Then the smallest positive integer for which P(n) is true is ______.


Prove the statement by using the Principle of Mathematical Induction:

For any natural number n, 7n – 2n is divisible by 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 for all n ∈ N.
cos α + cos(α + β) + cos(α + 2β) + ... + cos(α + (n – 1)β) = `(cos(alpha + ((n - 1)/2)beta)sin((nbeta)/2))/(sin  beta/2)`.


Prove that, sinθ + sin2θ + sin3θ + ... + sinnθ = `((sin ntheta)/2 sin  ((n + 1))/2 theta)/(sin  theta/2)`, for all n ∈ N.


Prove that `1/(n + 1) + 1/(n + 2) + ... + 1/(2n) > 13/24`, for all natural numbers n > 1.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×