मराठी

Prove that 1n+1+1n+2+...+12n>1324, for all natural numbers n > 1.

Advertisements
Advertisements

प्रश्न

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

सिद्धांत
Advertisements

उत्तर

Let P(n): `1/(n + 1) + 1/(n + 2) + ... + 1/(2n) > 13/24` ∀ n ∈ N

Step 1: P(2) : `1/(2 + 1) + 1/(2 + 2) > 13/24`

⇒ `1/3 + 1/4 > 13/24`

⇒ `7/12 > 13/24`

⇒ `14/24 > 13/24` which is true for P(2).

Step 2: P(k) : `1/(k + 1) + 1/(k + 2) + ... + 1/(2k) > 13/24`.

Let it be true for P(k).

Step 3: P(k + 1) : `1/(k + 1) + 1/(k + 2) + ... + 1/(2k) + 1/(2(k + 1)) > 13/24`

Since `1/(k + 1) + 1/(k + 2) + ... + 1/(2k) > 13/24`

So `1/(k + 1) + 1/(k + 2) + ... + 1/(2k) + 1/(2(k + 1)) > 13/24`

Which is true for P(k + 1).

Hence, P(k + 1) is true whenever P(k) is true.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Principle of Mathematical Induction - Exercise [पृष्ठ ७२]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
पाठ 4 Principle of Mathematical Induction
Exercise | Q 24 | पृष्ठ ७२

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

1.3 + 2.3^3 + 3.3^3  +...+ n.3^n = `((2n -1)3^(n+1) + 3)/4`

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

`1/2.5 + 1/5.8 + 1/8.11 + ... + 1/((3n - 1)(3n + 2)) = n/(6n + 4)`

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

`a + ar + ar^2 + ... + ar^(n -1) = (a(r^n - 1))/(r -1)`

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: 

`1/1.4 + 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 (2n +7) < (n + 3)2


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 + 2.3 + 3.4 + ... + n (n + 1) = \[\frac{n(n + 1)(n + 2)}{3}\]

 

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


32n+2 −8n − 9 is divisible by 8 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 ∈ N 

 


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

 


\[\text{ Prove that }  \frac{1}{n + 1} + \frac{1}{n + 2} + . . . + \frac{1}{2n} > \frac{13}{24}, \text{ for all natural numbers } n > 1 .\]

 


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


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


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

2 + 4 + 6 + ..... + 2n = n (n+1)


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

Given that tn+1 = 5tn + 4, t1 = 4, prove that tn = 5n − 1


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

22n – 1 is divisible by 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.


Prove by the Principle of Mathematical Induction that 1 × 1! + 2 × 2! + 3 × 3! + ... + n × n! = (n + 1)! – 1 for all natural numbers n.


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`


Prove the statement by using the Principle of Mathematical Induction:

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


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 the statement by using the Principle of Mathematical Induction:

n2 < 2n for all natural numbers n ≥ 5.


Prove the statement by using the Principle of Mathematical Induction:

1 + 5 + 9 + ... + (4n – 3) = n(2n – 1) for all natural numbers n.


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


Show that `n^5/5 + n^3/3 + (7n)/15` is a natural number for all n ∈ N.


If 10n + 3.4n+2 + k is divisible by 9 for all n ∈ N, then the least positive integral value of k is ______.


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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×