Prove by method of induction, for all n ∈ N: Given that tn+1 = 5tn + 4, t1 = 4, prove that tn = 5n − 1 - Mathematics and Statistics

प्रश्न

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

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

योग

उत्तर

The statement P(n) has L.H.S. a recurrence relation t_n+1 = 5t_n + 4, t₁ = 4 and R.H.S. a general statement t_n = 5ⁿ – 1

Step I:

To prove P(1) is true.

L.H.S. = 4, R.H.S. = 5¹ – 1 = 4

∴ P(1) is true.

For n = 2, L.H.S. = t₂ = 5t₁ + 4 = 24

R.H.S. = t₂ = 5² – 1 = 24

∴ P(2) is also true.

Step II:

Assume P(k) is true.

i.e. t_k+1 = 5t_k + 4 and t_k = 5^k – 1

Step III:

To prove P(k + 1) is true.

i.e. to prove t_k+1 = 5^k+1 – 1

Since t_k+1 = 5t_k + 4 and t_k = 5^k – 1 .........(From Step II)

∴ t_k+1 = 5(5^k – 1) + 4 = 5^k+1 – 1

∴ P(k + 1) is true.

From all the steps above,

P(n) : t_n = 5ⁿ – 1 is true for all

n ∈ N where t_n+1 = 5t_n + 4 , t₁ = 4

shaalaa.com

Principle of Mathematical Induction

क्या इस प्रश्न या उत्तर में कोई त्रुटि है?

Q 15 Q 14 Q 16

अध्याय 4: Methods of Induction and Binomial Theorem - Exercise 4.1 [पृष्ठ ७४]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board

अध्याय 4 Methods of Induction and Binomial Theorem
Exercise 4.1 | Q 15 | पृष्ठ ७४

संबंधित प्रश्न

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+ 1/1)(1+ 1/2)(1+ 1/3)...(1+ 1/n) = (n + 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))`

If P (n) is the statement "n(n + 1) is even", then what is P(3)?

If P (n) is the statement "n3 + n is divisible by 3", prove that P (3) is true but P (4) is not true.

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

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

1 + 3 + 5 + ... + (2n − 1) = n² i.e., the sum of first n odd natural numbers is n².

\[\frac{1}{3 . 5} + \frac{1}{5 . 7} + \frac{1}{7 . 9} + . . . + \frac{1}{(2n + 1)(2n + 3)} = \frac{n}{3(2n + 3)}\]

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

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

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

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)`

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

3ⁿ − 2n − 1 is divisible by 4

Answer the following:

Prove by method of induction log_a xⁿ = n log_ax, x > 0, n ∈ N

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

1 + 3 + 5 + ... + (2n – 1) = n²

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.

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`