हिंदी

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 .

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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 12: Mathematical Induction - Exercise 12.2 [पृष्ठ २९]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 12 Mathematical Induction
Exercise 12.2 | Q 48 | पृष्ठ २९

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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+ 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 + 2.3.4 + … + 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

`1/1.4 + 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

`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 (2+7) < (n + 3)2


If P (n) is the statement "n3 + n is divisible by 3", prove that P (3) is true but P (4) is not true.


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

 

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

 

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


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

 

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


a + (a + d) + (a + 2d) + ... (a + (n − 1) d) = \[\frac{n}{2}\left[ 2a + (n - 1)d \right]\]

 


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

 

(ab)n = anbn for all n ∈ N. 

 

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

 


\[\sin x + \sin 3x + . . . + \sin (2n - 1)x = \frac{\sin^2 nx}{\sin x}\]

 


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

 


\[\text { A sequence  } x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_1 = 2 \text{ and }  x_k = \frac{x_{k - 1}}{k} \text{ for all natural numbers } k, k \geq 2 . \text{ Show that }  x_n = \frac{2}{n!} \text{ for all } n \in N .\]


\[\text{ The distributive law from algebra states that for all real numbers}  c, a_1 \text{ and }  a_2 , \text{ we have }  c\left( a_1 + a_2 \right) = c a_1 + c a_2 . \]
\[\text{ Use this law and mathematical induction to prove that, for all natural numbers, } n \geq 2, if c, a_1 , a_2 , . . . , a_n \text{ are any real numbers, then } \]
\[c\left( a_1 + a_2 + . . . + a_n \right) = c a_1 + c a_2 + . . . + c a_n\]


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:

3n − 2n − 1 is divisible by 4


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

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


Answer the following:

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

12 + 42 + 72 + ... + (3n − 2)2 = `"n"/2 (6"n"^2 - 3"n" - 1)`


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


Answer the following:

Prove by method of induction 152n–1 + 1 is divisible by 16, for all n ∈ N.


Answer the following:

Prove by method of induction 52n − 22n is divisible by 3, for all n ∈ N


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

`(1 - 1/2^2).(1 - 1/3^2)...(1 - 1/n^2) = (n + 1)/(2n)`, for all natural numbers, n ≥ 2. 


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

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


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:

n(n2 + 5) is divisible by 6, for each natural number n.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×