हिंदी

Prove that Cos α + Cos ( α + β ) + Cos ( α + 2 β ) + . . . + Cos [ α + ( N − 1 ) β ] = Cos { α + ( N − 1 2 ) β } Sin ( N β 2 ) Sin ( β 2 ) for All N ∈ N - Mathematics

Advertisements
Advertisements

प्रश्न

\[\text{ Prove that } \cos\alpha + \cos\left( \alpha + \beta \right) + \cos\left( \alpha + 2\beta \right) + . . . + \cos\left[ \alpha + \left( n - 1 \right)\beta \right] = \frac{\cos\left\{ \alpha + \left( \frac{n - 1}{2} \right)\beta \right\}\sin\left( \frac{n\beta}{2} \right)}{\sin\left( \frac{\beta}{2} \right)} \text{ for all n } \in N .\]

 

उत्तर

\[\text{ Let p } \left( n \right): \cos\alpha + \cos\left( \alpha + \beta \right) + \cos\left( \alpha + 2\beta \right) + . . . + \cos\left[ \alpha + \left( n - 1 \right)\beta \right] = \frac{\cos\left\{ \alpha + \left( \frac{n - 1}{2} \right)\beta \right\}\sin\left( \frac{n\beta}{2} \right)}{\sin\left( \frac{\beta}{2} \right)} \forall n \in  N . \] 

\[\text{ Step I: For }  n = 1, \]

\[LHS = \cos\left[ \alpha + \left( 1 - 1 \right)\beta \right] = \cos\alpha\]

\[RHS = \frac{\cos\left\{ \alpha + \left( \frac{1 - 1}{2} \right)\beta \right\}\sin\left( \frac{\beta}{2} \right)}{\sin\left( \frac{\beta}{2} \right)} = \cos\alpha\]

\[\text{ As, LHS = RHS } \]

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

\[\text{ Step II: For n = k,}  \]

\[ \text{ Let } p\left( k \right): \cos\alpha + \cos\left( \alpha + \beta \right) + \cos\left( \alpha + 2\beta \right) + . . . + \cos\left[ \alpha + \left( k - 1 \right)\beta \right] = \frac{\cos\left\{ \alpha + \left( \frac{k - 1}{2} \right)\beta \right\}\sin\left( \frac{k\beta}{2} \right)}{\sin\left( \frac{\beta}{2} \right)} \text{ be true  } \forall k \in N . \]

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

\[LHS = \cos\alpha + \cos\left( \alpha + \beta \right) + \cos\left( \alpha + 2\beta \right) + . . . + \cos\left[ \alpha + \left( k - 1 \right)\beta \right] + \cos\left[ \alpha + \left( k + 1 - 1 \right)\beta \right]\]

\[ = \frac{\cos\left\{ \alpha + \left( \frac{k - 1}{2} \right)\beta \right\}\sin\left( \frac{k\beta}{2} \right)}{\sin\left( \frac{\beta}{2} \right)} + \cos\left( \alpha + k\beta \right)\]

\[ = \frac{\cos\left\{ \alpha + \left( \frac{k - 1}{2} \right)\beta \right\}\sin\left( \frac{k\beta}{2} \right) + \sin\left( \frac{\beta}{2} \right)\cos\left( \alpha + k\beta \right)}{\sin\left( \frac{\beta}{2} \right)}\]

\[ = \frac{\sin\left( \alpha + k\beta - \frac{\beta}{2} \right) - \sin\left( \alpha - \frac{\beta}{2} \right) + \sin\left( \alpha + k\beta + \frac{\beta}{2} \right) - \sin\left( \alpha + k\beta - \frac{\beta}{2} \right)}{2\sin\left( \frac{\beta}{2} \right)}\]

\[ = \frac{- \sin\left( \alpha - \frac{\beta}{2} \right) + \sin\left( \alpha + k\beta + \frac{\beta}{2} \right)}{2\sin\left( \frac{\beta}{2} \right)}\]

\[ = \frac{2\cos\left( \frac{2\alpha + k\beta}{2} \right)\sin\left( \frac{k\beta + \beta}{2} \right)}{2\sin\left( \frac{\beta}{2} \right)}\]

\[ = \frac{\cos\left( \alpha + \frac{k\beta}{2} \right)\sin\left( \frac{\left( k + 1 \right)\beta}{2} \right)}{\sin\left( \frac{\beta}{2} \right)}\]

\[RHS = \frac{\cos\left\{ \alpha + \left( \frac{k + 1 - 1}{2} \right)\beta \right\}\sin\left( \frac{\left( k + 1 \right)\beta}{2} \right)}{\sin\left( \frac{\beta}{2} \right)}\]

\[ = \frac{\cos\left( \alpha + \frac{k\beta}{2} \right)\sin\left( \frac{\left( k + 1 \right)\beta}{2} \right)}{\sin\left( \frac{\beta}{2} \right)}\]

\[As, LHS = RHS\]

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

\[\text{ Hence,}  \cos\alpha + \cos\left( \alpha + \beta \right) + \cos\left( \alpha + 2\beta \right) + . . . + \cos\left[ \alpha + \left( n - 1 \right)\beta \right] = \frac{\cos\left\{ \alpha + \left( \frac{n - 1}{2} \right)\beta \right\}\sin\left( \frac{n\beta}{2} \right)}{\sin\left( \frac{\beta}{2} \right)} \text{ for all } n \in N .\]

 

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

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 12 Mathematical Induction
Exercise 12.2 | Q 40 | 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: 1.2 + 2.22 + 3.22 + … + n.2n = (n – 1) 2n+1 + 2


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

`1/3.5 + 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 ∈ Nn (n + 1) (n + 5) is a multiple of 3.


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


\[\frac{1}{3 . 7} + \frac{1}{7 . 11} + \frac{1}{11 . 5} + . . . + \frac{1}{(4n - 1)(4n + 3)} = \frac{n}{3(4n + 3)}\] 


1.3 + 2.4 + 3.5 + ... + n. (n + 2) = \[\frac{1}{6}n(n + 1)(2n + 7)\]

 

1.2 + 2.3 + 3.4 + ... + n (n + 1) = \[\frac{n(n + 1)(n + 2)}{3}\]

 

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

 

n(n + 1) (n + 5) is a multiple of 3 for all n ∈ N.

 

Prove that 1 + 2 + 22 + ... + 2n = 2n+1 - 1 for all \[\in\] N .

 

Let P(n) be the statement : 2n ≥ 3n. If P(r) is true, show that P(r + 1) is true. Do you conclude that P(n) is true for all n ∈ N


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

 


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


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

 


\[\text{ A sequence }  a_1 , a_2 , a_3 , . . . \text{ is defined by letting }  a_1 = 3 \text{ and } a_k = 7 a_{k - 1} \text{ for all natural numbers } k \geq 2 . \text{ Show that } a_n = 3 \cdot 7^{n - 1} \text{ for all } n \in N .\]


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

1.3 + 3.5 + 5.7 + ..... to n terms = `"n"/3(4"n"^2 + 6"n" - 1)`


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:

Prove by method of induction loga xn = n logax, x > 0, 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:

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


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))`


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


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:

n3 – 7n + 3 is divisible by 3, for all natural numbers n.


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:

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


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


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×