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 = $\frac{n(4n^2+6n-1)}{3}$, 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) =$\frac{n(4n^2+6n-1)}{3}$

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)

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

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

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)

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

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