Question

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

Answer 1

Let p(n) : $1.3+3.5+5.7+....+(2n-1)(2n+1)=\frac{n(4n^2+6n-1)}{6}$

For n = 1

$1.3=\frac{1(4+6-1)}{3}$

$3=3$

$\Rightarrow$ P(n) is true for n = 1

Let P(n) is true for n = k, so

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

We have to show that,

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

Now,

$\{1.3+3.5+5.7+....+(2k-1\left)\right(2k+1\left)\right\}+(2k+1)(2k+3)$

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

[Using equation (1)]

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

$=\frac{4k^3+6k^2-k+12k^2+18k+6k+9}{3}$

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

$=\frac{4k^3+4k^2+14k^2+14k+9k+9}{3}$

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

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

$\Rightarrow$ p(n) is true for n = k + 1

$\Rightarrow$ p(n) is true for all n $epsilon$ N by PMI.