Prove by induction that for all natural number n sinα + sin(α + β) + sin(α + 2β)+ ... + sin(α + (n – 1)β) = sin(α+n-12β)sin(nβ2)sin(β2)

Question

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))`

Sum

Solution

Consider P(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))`, for all natural number n.

We observe that P(1) is true.

Since P(1): sin α = `(sin(alpha + 0) sin beta/2)/(sin beta/2)`

Assume that P(n) is true for some natural numbers k.

i.e., P(k): sin α + sin(α + β) + sin(α + 2β) + ... + sin(α + (k – 1)β)

= `(sin (alpha + (k - 1)/2 beta)sin((kbeta)/2))/(sin(beta/2))`

Now, to prove that P(k + 1) is true.

We have P(k + 1): sin α + sin(α + β) + sin(α + 2β) + ... + sin(α + (k – 1)β) + sin(α + kβ)

= `(sin (alpha + ("k" - 1)/2 beta)sin((kbeta)/2))/(sin(beta/2)) + sin(alpha + kbeta)`

= `(sin(alpha + (k - 1)/2 beta) sin (kbeta)/2 + sin(alpha + kbeta) sin beta/2)/(sin beta/2)`

= `(cos(alpha - beta/2) - cos(alpha + kbeta - beta/2) + cos(alpha + kbeta - beta/2) - cos(alpha + kbeta + beta/2))/(2sin beta/2)`

= `(cos(alpha - beta/2) - cos(alpha + kbeta + beta/2))/(2sin beta/2)`

= `(sin (alpha + (kbeta)/2)sin ((kbeta + beta)/2))/(sin beta/2)`

= `(sin(alpha + (kbeta)/2) sin(k + 1)(beta/2))/(sin beta/2)`

Thus P(k + 1) is true whenever P(k) is true.

Hence, by the Principle of Mathematical Induction P(n) is true for all natural number n.

shaalaa.com

Principle of Mathematical Induction

Is there an error in this question or solution?

Q 8 Q 7 Q 9

Chapter 4: Principle of Mathematical Induction - Solved Examples [Page 65]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 11

Chapter 4 Principle of Mathematical Induction
Solved Examples | Q 8 | Page 65

Video TutorialsVIEW ALL [1]

view
Video Tutorials For All Subjects
Principle of Mathematical Induction

video tutorial00:17:42

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:

`1/2.5 + 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:

`1^2 + 3^2 + 5^2 + ... + (2n -1)^2 = (n(2n - 1) (2n + 1))/3`

Given an example of a statement P (n) such that it is true for all n ∈ N.

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.

1 + 2 + 3 + ... + n = \[\frac{n(n + 1)}{2}\] i.e. the sum of the first n natural numbers is \[\frac{n(n + 1)}{2}\] .

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

1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) =\[\frac{n(4 n^2 + 6n - 1)}{3}\]

5²ⁿ⁺² −24n −25 is divisible by 576 for all n ∈ N.

3²ⁿ⁺² −8n − 9 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 ∈ N

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

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

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

Prove that the number of subsets of a set containing n distinct elements is 2ⁿ, 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:

`1/(1.3) + 1/(3.5) + 1/(5.7) + ... + 1/((2"n" - 1)(2"n" + 1)) = "n"/(2"n" + 1)`

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

(2³ⁿ − 1) is divisible by 7

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

(2⁴ⁿ−1) is divisible by 15

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

3ⁿ − 2n − 1 is divisible by 4

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.

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, ..., a_n are any real numbers, then c(a₁ + a₂ + ... + a_n) = ca₁ + ca₂ + ... + ca_n.