मराठी

Prove that, cosθ cos2θ cos22θ ... cos2n–1θ = sin2nθ2nsinθ, for all n ∈ N.

Advertisements
Advertisements

प्रश्न

Prove that, cosθ cos2θ cos22θ ... cos2n–1θ = `(sin 2^n theta)/(2^n sin theta)`, for all n ∈ N.

सिद्धांत
Advertisements

उत्तर

Let P(n): cosθ cos2θ cos22θ ... cos2n–1θ = `(sin 2^n theta)/(2^n sin theta)`, ∀ n ∈ N.

Step 1: P(1): cosθ = `(sin 2^1 theta)/(2^1 sin theta)`

= `(sin 2theta)/(2sin theta)`

= `(2sin theta cos theta)/(2sin theta)`

= cosθ

⇒ cosθ = cosθ which is true for P(1)

Step 2: P(k): cosθ.cos2θ.cos22θ ... cos2k – 1θ = `(sin 2^k theta)/(2^k sin theta)`

Let it be true for P(k).

Step 3: P(k + 1): cosθ.cos2θ.cos22θ ... cos2k – 1θ . cos`2^(("k"+1)–1`θ

= `(sin 2^k theta)/(2^k sin theta) . cos 2^((k + 1) - 1)theta`

=  `(sin 2^k theta)/(2^k sin theta) . cos 2^ktheta`

= `(2 sin 2^k theta . cos 2^k theta)/(2.2^k sin theta)`

= `(sin 2.2^k theta)/(2^(k + 1) sin theta)`  .....[∵ 2 sinθ cosθ = sin2θ]

= `(sin 2^(k + 1)theta)/(2^(k + 1) sin theta)` 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 ७१]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
पाठ 4 Principle of Mathematical Induction
Exercise | Q 21 | पृष्ठ ७१

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

`1^2 + 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: 41n – 14n is a multiple of 27.


If P (n) is the statement "n2 − n + 41 is prime", prove that P (1), P (2) and P (3) are true. Prove also that P (41) is not true.


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

 

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

 

(ab)n = anbn for all n ∈ N. 

 

72n + 23n−3. 3n−1 is divisible by 25 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 .

  

\[\frac{n^7}{7} + \frac{n^5}{5} + \frac{n^3}{3} + \frac{n^2}{2} - \frac{37}{210}n\] is a positive integer for all n ∈ N.  

 


\[\frac{1}{2}\tan\left( \frac{x}{2} \right) + \frac{1}{4}\tan\left( \frac{x}{4} \right) + . . . + \frac{1}{2^n}\tan\left( \frac{x}{2^n} \right) = \frac{1}{2^n}\cot\left( \frac{x}{2^n} \right) - \cot x\] for all n ∈ and  \[0 < x < \frac{\pi}{2}\]

 


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


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


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


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


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:

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


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:

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

2 + 3.2 + 4.22 + ... + (n + 1)2n–1 = n.2n 


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 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 + 3 + 5 + ... + (2n – 1) = n2 


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

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


Prove by induction that for all natural number n sinα + sin(α + β) + sin(α + 2β)+ ... + sin(α + (n – 1)β) = `(sin (alpha + (n - 1)/2 beta)sin((nbeta)/2))/(sin(beta/2))`


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"):}`


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


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:

`sqrt(n) < 1/sqrt(1) + 1/sqrt(2) + ... + 1/sqrt(n)`, for all natural numbers n ≥ 2.


Prove the statement by using the Principle of Mathematical Induction:

2 + 4 + 6 + ... + 2n = n2 + n for all natural numbers n.


A sequence b0, b1, b2 ... is defined by letting b0 = 5 and bk = 4 + bk – 1 for all natural numbers k. Show that bn = 5 + 4n for all natural number n using mathematical induction.


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


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


If P(n): 2n < n!, n ∈ N, then P(n) is true for all n ≥ ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×