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