Introduction to P.M.I
Question

# 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

Moderate
Solution

## $\because {\mathrm{u}}_{\mathrm{n}+1}=3{\mathrm{u}}_{\mathrm{n}}-2{\mathrm{u}}_{\mathrm{n}-1}----\left(\mathrm{i}\right)$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 = 2step 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 getThis shows that the result is true for n =k + 1, hence by the principle of mathematical induction the result is true for all $\mathrm{n}\in \mathrm{N}$

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