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Messagees are transmitted over a communication channel using two signals. the transmission of one signal requires 1 microsecond and the transmission of the other requires two microseconds. the recurrence relation for the numver of different massages consisting of sequences of these two signals ( where each signal is immediate followed by the next signal ) that can be send in ‘n’ microseconds (n>=2) is

1. $a_{n} = a_{n-1} + a_{n-2}$
2. $a_{n} = a_{n-1} + 2a_{n-2}$
3. $a_{n} = 2a_{n-1} + a_{n-2}$
4. $a_{n} = a_{n-1} + a_{n-3}$
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$a_n$ is the number of different messages that can be sent in exactly $n$ microseconds.

$\implies a_{n-1}=$ number of different messages that can be sent in $n-1$ microseconds,

and $a_{n-2}=$ number of different messages that can be sent in $n-2$ microseconds.

### For the $n^{th}$ microsecond:

If the last signal is of $1$ microsecond:

Let the messages possible in this case be $end_1$. Now, in this case you can merge last signal with any message that was possible in $n-1$ microseconds, hence you can have $a_{n-1}$ different messages which will have last signal of $1$ microsecond.

$\therefore end_1=a_{n-1}$

If the last signal is of $2$ microseconds:

Let the messages possible in this case be $end_2$. In this case you can merge last signal with any message that was possible in $n-2$ microseconds, hence you can have $a_{n-2}$ different messages which will have last signal of $2$ microseconds.

$\therefore end_2=a_{n-2}$

Since the last signal can be of either $1$ microsecond or $2$ microsecond, the total number of different messages that can be sent in $n$ microseconds , i.e. $a_n$ will be:

$end_1 + end_2$

$=a_{n-1}+a_{n-2}$

Hence, option $1.\ a_n=a_{n-1}+a_{n-2}$ is correct.

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