If ${\mathrm{u}}_{\mathrm{n}+1}=3{\mathrm{u}}_{\mathrm{n}}-2{\mathrm{u}}_{\mathrm{n}-1}$ and ${\mathrm{u}}_0=2,{\mathrm{u}}_1=3$ then ${\mathrm{u}}_{\mathrm{n}}$ s equal to

$\because {\mathrm{u}}_{\mathrm{n}+1}=3{\mathrm{u}}_{\mathrm{n}}-2{\mathrm{u}}_{\mathrm{n}-1}----\left(\mathrm{i}\right)$

step I : Given ${\mathrm{u}}_1=3=2+1=2^1+1$ which is true for n =1 , 

Put n =1in Eq. (i), 

$\begin{array}{c}{\mathrm{u}}_{1+1}=3{\mathrm{u}}_1-2{\mathrm{u}}_{1-1}\\ {\mathrm{u}}_2=3{\mathrm{u}}_1-2{\mathrm{u}}_0\\ =3\cdot 3-2\cdot 2=5\\ =2^2+1\end{array}$

which is true for n =2. Therefore, the result are true for n = 1 and n = 2

step II : Assume it is true for n = &, then it is also true for n =k-1.

  $\begin{array}{l}{\mathrm{u}}_{\mathrm{k}}=2^{\mathrm{k}}+1---\left(\mathrm{i}\right)\\ {\mathrm{u}}_{\mathrm{k}-1}=2^{\mathrm{k}-1}+1---\left(\mathrm{ii}\right)\end{array}$

Step lll : On putting n = k in Eq. (i), we get

         $\begin{array}{c}{\mathrm{u}}_{\mathrm{k}+1}=3{\mathrm{u}}_{\mathrm{k}}-2{\mathrm{u}}_{\mathrm{k}-1}\\ =3\left(2^{\mathrm{k}}+1\right)-2\left(2^{\mathrm{k}-1}+1\right)\text{ [from Eqs. (ii) and (iii)] }\\ =3\cdot 2^{\mathrm{k}}+3-2\cdot 2^{\mathrm{k}-1}-2\\ =3\cdot 2^{\mathrm{k}}+3-2^{\mathrm{k}}-2\\ =(3-1)2^{\mathrm{k}}+1\\ =2\cdot 2^{\mathrm{k}}+1\\ =2^{\mathrm{k}+1}+1\end{array}$

This shows that the result is true for n =k + 1, hence by the principle of mathematical induction the result is true for all $\mathrm{n}\in \mathrm{N}$