Prove the following by using the principle of mathematical induction for all $n in N: 1.3+3.5+5.7+ldots+(2 n-1)(2 n+1)=$

Question:

Prove the following by using the principle of mathematical induction for all n ∈ N$1.3+3.5+5.7+\ldots+(2 n-1)(2 n+1)=\frac{n\left(4 n^{2}+6 n-1\right)}{3}$

Solution:

Let the given statement be P(n), i.e.,

$\mathrm{P}(n): 1.3+3.5+5.7+\ldots+(2 n-1)(2 n+1)=\frac{n\left(4 n^{2}+6 n-1\right)}{3}$

For n = 1, we have

$P(1): 1.3=3=\frac{1\left(4.1^{2}+6.1-1\right)}{3}=\frac{4+6-1}{3}=\frac{9}{3}=3$, which is true.

Let P(k) be true for some positive integer k, i.e.,

$1.3+3.5+5.7+\ldots \ldots+(2 k-1)(2 k+1)=\frac{k\left(4 k^{2}+6 k-1\right)}{3}$  $\ldots$ (i)

We shall now prove that P(k + 1) is true.

Consider

$(1.3+3.5+5.7+\ldots+(2 k-1)(2 k+1)+\{2(k+1)-1\} 2(k+1)+1\}$

$=\frac{k\left(4 k^{2}+6 k-1\right)}{3}+(2 k+2-1)(2 k+2+1)$ [Using (i)]

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

$=\frac{k\left(4 k^{2}+6 k-1\right)}{3}+\left(4 k^{2}+8 k+3\right)$

$=\frac{k\left(4 k^{2}+6 k-1\right)+3\left(4 k^{2}+8 k+3\right)}{3}$

$=\frac{4 k^{3}+6 k^{2}-k+12 k^{2}+24 k+9}{3}$

$=\frac{4 k^{3}+18 k^{2}+23 k+9}{3}$

$=\frac{k\left(4 k^{2}+14 k+9\right)+1\left(4 k^{2}+14 k+9\right)}{3}$

$=\frac{(k+1)\left(4 k^{2}+14 k+9\right)}{3}$

$=\frac{(k+1)\left\{4 k^{2}+8 k+4+6 k+6-1\right\}}{3}$

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

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

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.