First slide

Introduction to P.M.I

Question

If $u_{n + 1} = 3 u_{n} - 2 u_{n - 1}$ and $u_{0} = 2, u_{1} = 3$ then $u_{n}$ s equal to

Moderate

A

$1 - 2^{n}$
B

$2^{n} + 1$
C

$2^{n} - 1$
D

$2^{n} + 2$

Solution

$∵ u_{n + 1} = 3 u_{n} - 2 u_{n - 1} - - - - (i)$
step I : Given $u_{1} = 3 = 2 + 1 = 2^{1} + 1$ which is true for n =1 ,
Put n =1in Eq. (i),
$\begin{matrix} u_{1 + 1} = 3 u_{1} - 2 u_{1 - 1} \\ u_{2} = 3 u_{1} - 2 u_{0} \\ = 3 \cdot 3 - 2 \cdot 2 = 5 \\ = 2^{2} + 1 \end{matrix}$
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.
$\begin{array}{l} u_{k} = 2^{k} + 1 - - - (i) \\ u_{k - 1} = 2^{k - 1} + 1 - - - (ii) \end{array}$
Step lll : On putting n = k in Eq. (i), we get
$\begin{matrix} u_{k + 1} = 3 u_{k} - 2 u_{k - 1} \\ = 3 (2^{k} + 1) - 2 (2^{k - 1} + 1) [from Eqs. (ii) and (iii)] \\ = 3 \cdot 2^{k} + 3 - 2 \cdot 2^{k - 1} - 2 \\ = 3 \cdot 2^{k} + 3 - 2^{k} - 2 \\ = (3 - 1) 2^{k} + 1 \\ = 2 \cdot 2^{k} + 1 \\ = 2^{k + 1} + 1 \end{matrix}$
This shows that the result is true for n =k + 1, hence by the principle of mathematical induction the result is true for all $n \in N$

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