Prove by method of induction, 1.3+3.5+5.7+...........+(2n−1)(2n+1)=$\frac{{n(4{n^2} + 6n - 1)}}{3}$

Solution:

Let P(n) ≡ 1.3 + 3.5 + 5.7 + ..... to n terms = $\frac{{n(4{n^2} + 6n - 1)}}{3}$, for all n ∈ N

But the first factor in each term

i.e., 1, 3, 5 … are in A.P. with a = 1 and d = 2.

∴ n^th 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.

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

∴ n^th term, t_n = (2n – 1) (2n + 1)

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

Step I:

Put n = 1

L.H.S. = 1.3 = 3

R.H.S. =$\frac{{1(4{{(1)}^2} + 6(1) - 1)}}{3} = 3$=L.H.S.

∴ P(n) is true for n = 1.

Step II:

Let us consider that P(n) is true for n = k

∴ 1.3 + 3.5 + 5.7 + ..... + (2k – 1)(2k + 1)

=$\frac{{k(4{k^2} + 6k - 1)}}{3}$....(i)

Step III:

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

i.e., to prove that

1.3 + 3.5 + 5.7 + …. + [2(k + 1) – 1][2(k + 1) + 1]

$\begin{array}{l} = \frac{{(k + 1)[4{{(k + 1)}^2} + 6(k + 1) - 1]}}{3}\\ = \frac{{(k + 1)[4{k^2} + 8k + 4 + 6k + 6 - 1]}}{3}\\ = \frac{{(k + 1)}}{3}(4{k^2} + 14k + 9)\end{array}$

L.H.S. = 1.3 + 3.5 + 5.7 + ... + [2(k + 1) – 1][2(k + 1) + 1]

= 1.3 + 3.5 + 5.7 + ... + (2k – 1)(2k + 1) + (2k + 1)(2k + 3)

 $\begin{array}{l} = \frac{k}{3}(4{k^2} + 6k - 1) + (2k + 1)(2k + 3).......[From{\rm{ (i)}}]\\ = \frac{1}{3}[(4{k^3} + 6{k^2} - k) + 3(2k + 1)(2k + 3)]\\ = \frac{1}{3}\left( {4{k^3} + 6{k^2} - k + 12{k^2} + 24k + 9} \right)\\ = \frac{1}{3}\left( {4{k^3} + 18{k^2} + 23k + 9} \right)\\ = \frac{1}{3}(k + 1)(4{k^2} + 14k + 9)\end{array}$

= R.H.S.

∴ P(n) is true for n = k + 1

Step IV:

From all 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 =$\frac{{n(4{n^2} + 6n - 1)}}{3}$ for all n ∈ N.