Question

Given $U_{n + 1} = 3U_{n} - 2U_{n - 1}$ and $U_{0} = 2$, $U_{1} = 3$, the value of $U_{n}$ for all $n \in N$ is

• A. $2^{n} - 1$
• B. $2^{n} - 1$
• C. 0
• D. None of these

Answer 1

Answer: (B)

$2^{n} - 1$

Answer 2

$\because U_{n + 1} = 3U_{n} - 2U_{n - 1}$ …..(i)

Step I : Given $U_{1} = 3$

For n =1, $U_{1 + 1} = 3U_{1} - 2U_{0}$, $U_{2} = 3.3 - 2.2 = 5$

Option (2) $U_{n} = 2^{n} + 1$

For n = 1, $U_{1} = 2^{1} + 1 = 3$ which is true. For n = 2, $U_{2} = 2^{2} + 1 = 5$ which is true

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

Step II : Assume it is true for n = k then it is also true for n = k – 1

Then $U_{k} = 2^{k} + 1$ …..(ii) and $U_{k - 1} = 2^{k - 1} + 1$…..(iii)

Step III : Putting n = k in (i), we get

$U_{k + 1} = 3U_{k} - 2{U_{k}}_{- 1} = 3\lbrack 2^{k} + 1\rbrack - 2\lbrack 2^{k - 1} + 1\rbrack = 3.2^{k} + 3 - 2.2^{k - 1} - 2 = 3.2^{k} + 1 - 2.2^{k - 1}$

$\Rightarrow 3.2^{k} - 2^{k} + 1 = 2.2^{k} + 1 = 2^{k + 1} + 1$

$\Rightarrow$ $U_{k + 1} = 2^{k + 1} + 1$

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