English

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

Advertisements
Advertisements

Question

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

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

Sum
Advertisements

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

∴ nth 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
  Is there an error in this question or solution?
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 +  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

1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ...+ `1/(n(n+1)(n+2)) = (n(n+3))/(4(n+1) (n+2))`

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 ∈ N: 102n – 1 + 1 is divisible by 11


Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 – 8n– 9 is divisible by 8.


If P (n) is the statement "n2 + n is even", and if P (r) is true, then P (r + 1) is true.

 

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.


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


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


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

 

2 + 5 + 8 + 11 + ... + (3n − 1) = \[\frac{1}{2}n(3n + 1)\]

 

a + (a + d) + (a + 2d) + ... (a + (n − 1) d) = \[\frac{n}{2}\left[ 2a + (n - 1)d \right]\]

 


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

 

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

 

Prove that n3 - 7+ 3 is divisible by 3 for all n \[\in\] 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.  

 


\[1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + . . . + \frac{1}{n^2} < 2 - \frac{1}{n}\] for all n ≥ 2, n ∈ 

 


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

 

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

 


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

2 + 4 + 6 + ..... + 2n = n (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:

(23n − 1) is divisible by 7


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

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


Answer the following:

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


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 by the Principle of Mathematical Induction that 1 × 1! + 2 × 2! + 3 × 3! + ... + n × n! = (n + 1)! – 1 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:

n2 < 2n for all natural numbers n ≥ 5.


Prove the statement by using the Principle of Mathematical Induction:

2n < (n + 2)! for all natural number n.


If 10n + 3.4n+2 + k is divisible by 9 for all n ∈ N, then the least positive integral value of k is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×