Answer:
a_{n} = 2/3 . 2^n + 1/3 . (-1)^n
Explanation:
Let a_{n} represents number of the message that can transmitted in <em>n </em>microsecond using three of different signals.
One signal requires one microsecond for transmittal: a_{n}-1
Another signal requires two microseconds for transmittal: a_{n}-2
The last signal requires two microseconds for transmittal: a_{n}-2
a_{n}= a_{n-1} + a_{n-2} + a_{n-2} = a_{n-1} + 2a_{n-2}, n ≥ 2
In 0 microseconds. exactly 1 message can be sent: the empty message.
a_{0}= 1
In 1 microsecond. exactly 1 message can be sent (using the one signal of one microseconds:
a_{0}= 1
2- Roots Characteristic equation
Let a_{n} = r^2, a_{n-1}=r and a_{n-2}= 1
r^2 = r+2
r^2 - r - 2 =0 Subtract r+6 from each side
(r - 2)(n+1)=0 Factorize
r - 2 = 0 or r +1 = 0 Zero product property
r = 2 or r = -1 Solve each equation
Solution recurrence relation
The solution of the recurrence relation is then of the form a_{1} = a_{1 r^n 1} + a_{2 r^n 2} with r_{1} and r_{2} the roots of the characteristic equation.
a_{n} =a_{1} . 2^n + a_{2}.(-1)"
Initial conditions
:
1 = a_{0} = a_{1} + a_{2}
1 = a_{1} = 2a_{1} - a_{2}
Add the previous two equations
2 = 3a_{1}
2/3 = a_{1}
Determine a_{2} from 1 = a_{1} + a_{2} and a_{1} = 2/3
a_{2} = 1 - a_{1} = 1 - 2 / 3 = 1/3
Thus, the solution of recursion relation is a_{n} = 2/3 . 2^n + 1/3 . (-1)^n
