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.

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Principle of Mathematical Induction

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

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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

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