Prove by method of induction, for all n ∈ N: 3 + 7 + 11 + ..... + to n terms = n(2n+1)

Question

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

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

Sum

Solution

Let P(n) ≡ 3 + 7 + 11 + ... to n terms = n(2n + 1), for all n ∈ N.

3, 7, 11, ... are in A.P. with a = 3, d = 4

∴ n^th term = a+ (n – 1)d = 3 + (n – 1)4 = 4n – 1

∴ P(n) ≡ 3 + 7 + 11 + ... + (4n – 1) = n (2n + 1)

Step 1: For n = 1, L.H.S. = 3

R.H.S. = 1(2 × 1 + 1) = 3

∴ L.H.S. = R.H.S. for n = 1

∴ P(1) is true.

Step 2: Let us assume that for some k ∈ N, P(k) is true, i.e., 3 + 7 + 11 + ... + (4k – 1) = k(2k + 1) ...(1)

Step 3: To prove that P(k + 1) is true, i.e., to prove that 3 + 7 + 11 + ... + (4k – 1) + (4k + 3) = (k + 1)(2k + 3)

Now, L.H.S. = 3 + 7 + 11 + ... + (4k – 1) + (4k + 3)

= k(2k + 1) + (4k + 3) ... [By (1)]

= 2k² + k + 4k + 3

= 2k² + 3k + 2k + 3

= k(2k + 3) + 1(2k + 3)

= (k + 1)(2k + 3)

= R.H.S

∴ P(k + 1) is true.

Step 4: From all the above steps and by the principle of mathematical induction, the result P(n) is true for all n ∈ N, i.e., 3 + 7 + 11 + ... to n terms = n(2n + 1), for all n ∈ N.

shaalaa.com

Principle of Mathematical Induction

Is there an error in this question or solution?

Q 2 Q 1 Q 3

Chapter 4: Methods of Induction and Binomial Theorem - Exercise 4.1 [Page 73]

APPEARS IN

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

Chapter 4 Methods of Induction and Binomial Theorem
Exercise 4.1 | Q 2 | Page 73

RELATED QUESTIONS

Prove the following by using the principle of mathematical induction for all n ∈ N:

`1 + 3 + 3^2 + ... + 3^(n – 1) =((3^n -1))/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+3/1)(1+ 5/4)(1+7/9)...`(1 + ((2n + 1))/n^2) = (n + 1)^2`

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: x²ⁿ – y²ⁿ is divisible by x + y.

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 is even", and if P (r) is true, then P (r + 1) is true.

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.

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

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

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

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

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

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

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

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

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

\[\text{ Using principle of mathematical induction, prove that } \sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + . . . + \frac{1}{\sqrt{n}} \text{ for all natural numbers } n \geq 2 .\]

\[\text{ The distributive law from algebra states that for all real numbers} c, a_1 \text{ and } a_2 , \text{ we have } c\left( a_1 + a_2 \right) = c a_1 + c a_2 . \]
\[\text{ Use this law and mathematical induction to prove that, for all natural numbers, } n \geq 2, if c, a_1 , a_2 , . . . , a_n \text{ are any real numbers, then } \]
\[c\left( a_1 + a_2 + . . . + a_n \right) = c a_1 + c a_2 + . . . + c a_n\]

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:

1.2 + 2.3 + 3.4 + ..... + n(n + 1) = `"n"/3 ("n" + 1)("n" + 2)`

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

Answer the following:

Given that t_n+1 = 5t_n − 8, t₁ = 3, prove by method of induction that t_n = 5ⁿ⁻¹ + 2

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

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

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

2²ⁿ – 1 is divisible by 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 S_n of the n term of the series 1² + 2 × 2² + 3² + 2 × 4² + 5² + 2 × 6² ... is given by

S_n = `{{:((n(n + 1)^2)/2",", "if n is even"),((n^2(n + 1))/2",", "if n is odd"):}`

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 ≥ 4 but P(1), P(2) and P(3) are not true. Justify your answer