Question

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

Answer 1

Let P(n) be the given statement.
Now,
$$\begin{gathered} P\left(n\right)=1.3+3.5+5.7+...+(2n-1)(2n+1)=\frac{n(4n^2+6n-1)}{3} \\ \mathrm{Step}1: \\ P\left(1\right)=1.3=3=\frac{1(4\times {\left(1\right)}^2+6\times 1-1)}{3} \\ \mathrm{Hence},P\left(1\right)\mathrm{is}\mathrm{true}. \\ \mathrm{Step}2: \\ \mathrm{Let}P\left(m\right)be\mathrm{true}. \\ \mathrm{Then}, \\ 1.3+3.5+...+(2m-1)(2m+1)=\frac{m(4m^2+6m-1)}{3} \\ To\mathrm{prove}:P(m+1)\mathrm{is}\mathrm{true}. \\ Thatis, \\ 1.3+3.5+...+(2m+1)(2m+3)=\frac{(m+1)\left[4(m+1{)}^2+6\left(m+1\right)-1\right]}{3} \\ \mathrm{Now},P\left(m\right)\mathrm{is}\mathrm{equal}\mathrm{to}: \\ 1.3+3.5+...+(2m-1)(2m+1)=\frac{m(4m^2+6m-1)}{3} \\ \Rightarrow 1.3+3.5+...+(2m-1)(2m+1)+(2m+1)(2m+3)=\frac{m(4m^2+6m-1)}{3}+(2m+1)(2m+3)\left[\mathrm{Adding}(2m+1)(2m+3)\mathrm{to}\mathrm{both}\mathrm{sides}\right] \\ \Rightarrow P(m+1)=\frac{m(4m^2+6m-1)+3(4m^2+8m+3)}{3} \\ \Rightarrow P(m+1)=\frac{4m^3+6m^2-m+12m^2+24m+9}{3}=\frac{4m^3+18m^2+23m+9}{3} \\ \Rightarrow P(m+1)=\frac{4m(m^2+2m+1)+10m^2+19m+9}{3} \\ =\frac{4m(m+1{)}^2+(10m+9\left)\right(m+1)}{3} \\ =\frac{(m+1)\left[4m(m+1)+10m+9\right]}{3} \\ =\frac{(m+1)}{3}(4m^2+8m+4+6m+5) \\ =\frac{(m+1)\left[4(m+1{)}^2+6\left(m+1\right)-1\right]}{3} \\ \mathrm{Thus},P(m+1)\mathrm{is}\mathrm{true}. \\ Bythep\mathrm{rinciple}\mathrm{of}m\mathrm{athematical}i\mathrm{nduction},P\left(n\right)\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{a}lln\in N. \end{gathered}$$