English

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

Advertisements
Advertisements

Question

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

 

Advertisements

Solution

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

\[\text{ Step I: For }  n = 1, \]

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

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

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

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

\[\text{ Step II: For n = k,}  \]

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

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

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

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

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

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

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

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

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

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

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

\[As, LHS = RHS\]

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

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

 

shaalaa.com
  Is there an error in this question or solution?
Chapter 12: Mathematical Induction - Exercise 12.2 [Page 29]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 12 Mathematical Induction
Exercise 12.2 | Q 40 | Page 29

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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 + 2.3 + 3.4+ ... + n(n+1) = `[(n(n+1)(n+2))/3]`


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 + 1/4 + 1/8 + ... + 1/2^n = 1 - 1/2^n`

 

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: 

`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: 41n – 14n is a multiple of 27.


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


12 + 32 + 52 + ... + (2n − 1)2 = \[\frac{1}{3}n(4 n^2 - 1)\]

 

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


32n+2 −8n − 9 is divisible by 8 for all n ∈ N.


\[\frac{n^{11}}{11} + \frac{n^5}{5} + \frac{n^3}{3} + \frac{62}{165}n\] 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 ∈ N and  \[0 < x < \frac{\pi}{2}\]

 


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


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

 

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


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

3 + 7 + 11 + ..... + to n terms = n(2n+1)


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:

13 + 33 + 53 + .... to n terms = n2(2n2 − 1)


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

(cos θ + i sin θ)n = cos (nθ) + i sin (nθ)


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

`[(1, 2),(0, 1)]^"n" = [(1, 2"n"),(0, 1)]` ∀ n ∈ N


Answer the following:

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

22n – 1 is divisible by 3.


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

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


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:

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


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:

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


A sequence d1, d2, d3 ... is defined by letting d1 = 2 and dk = `(d_(k - 1))/"k"` for all natural numbers, k ≥ 2. Show that dn = `2/(n!)` for all n ∈ N.


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


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


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


If xn – 1 is divisible by x – k, then the least positive integral value of k is ______.


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.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×