Prove by the method of induction, for all n ∈ N. 1.3 + 3.5 + 5.7 +… to n terms = n/3(4n^2 + 6n – 1)

Question

Prove by the method of induction, for all n ∈ N.

1.3 + 3.5 + 5.7 +… to n terms = \(\frac{n}3\)(4n² + 6n – 1)

Answer 1

Let P(n) = 1.3 + 3.5 + 5.7 +… to n terms = \(\frac{n}3\)(4n² + 6n -1), for all n ∈ N.

But first factor in each term, i.e., 1, 3, 5,… are in A.P. with a = 1 and d = 2.

∴ nth term = a + (n – 1)d = 1 + (n – 1) 2 = (2n – 1) Also, second factor in each term, i.e., 3, 5, 7, … are in A.P. with a = 3 and d = 2.

∴ nth term = a + (n – 1) d = 3 + (n – 1) 2 = (2n + 1)

∴ nth term, t_n = (2n – 1) (2n + 1)

∴ P(n) ≡ 1.3 + 3.5 + 5.7 + …. + (2n – 1) (2n + 1) = \(\frac{n}3\)(4n² + 6n – 1)

Step I: Put n = 1

L.H.S. = 1.3 = 3

R.H.S. = \(\frac13\)[4(1)² + 6(1) – 1] = 3

∴ L.H.S. = R.H.S.

∴ P(n) is term for n = 1.

Step II: Let us assume that P(n) is true for n = k.

∴ 1.3 + 3.5 + 5.7 +….+ (2k – 1)(2k + 1) = \(\frac{k}3\)(4k² + 6k – 1) ……(i)

Step III: We have to prove that P(n) is true for n = k + 1,

i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV: From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.

∴ 1.3 + 3.5 + 5.7 +… to n terms = \(\frac{n}3\)(4n² + 6n – 1) for all n ∈ N.