Question

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

Answer 1

Let P (n) = $1.3+3.5+5.7+\cdots +(2n-1)(2n+1)=\frac{n(4n^2+6n-1)}{3}.$
$\text{For}=(2\times 1-1)(2\times 1+1)=\frac{1\left[4\right(1{)}^2+6\times 1-1]}{3}\Rightarrow 1\times 3=\frac{9}{3}\Rightarrow 3=3\therefore P\left(1\right)\text{is true}$
Let P (n) be true for n = k
$\therefore P\left(k\right)=1.3+3.5+5.7+\cdots +(2k-1)(2k+1)=\frac{k(4k^2+6k-1)}{3}\text{For}n=k+1P(k+1)=1.3+3.5+5.7+\cdots +(2k-1)(2k+1)+\left[2\right(k+1)-1]\left[2\right(ki+1)+1]\therefore P(k+1)=\frac{k(4k62+6k-1)}{3}+(2k+1)(2k+3)=\frac{4k^3+6k^2-k+3(4k^2+8k+9)}{3}=\frac{4k^3+6k^2-k+12k^2+24k+9}{3}=\frac{4k^3+18k^2+23k+9}{3}=\frac{(k+1)(4k^2+14k+9)}{3}$
$\therefore$ P (k + 1) is true
Thus P (k) is true $\Rightarrow (k+1)$ is true hence by principle f amthematical induction

P (n) is true for all $nϵN.$