Find the solution to the following recurrence:
an=3an−1+7 for n2 with initial conditions a0=7
An = 3 An-1 + 7
A0 = 7
A1 = 3*7 + 7 = 28
A2 = 3 * 28 + 7 = 91
To find the solution to the given recurrence relation, we can use the method of forward substitution. Here's how you can do it:
1. Write down the initial condition(s): a0 = 7.
2. Substitute n = 1 into the recurrence relation: a1 = 3a0 + 7.
Substituting the value of a0, we get a1 = 3(7) + 7 = 28.
3. Substitute n = 2 into the recurrence relation: a2 = 3a1 + 7.
Substituting the value of a1, we get a2 = 3(28) + 7 = 91.
4. Repeat this process for higher values of n until you have reached the desired value.
Using this method, you can calculate a3, a4, a5, and so on, until you find the value of "an" for n = 2.
Here's the calculated values for a few iterations:
a1 = 28
a2 = 91
a3 = 280
a4 = 847
a5 = 2540
It seems that the solution to the given recurrence relation is a geometric sequence with a common ratio of 3 and an initial term of 7. So, the general formula for an using the initial condition a0 = 7 can be written as:
an = 7 x 3^(n-1)
Therefore, for n = 2, the solution to the recurrence relation is:
a2 = 7 x 3^(2-1) = 7 x 3^1 = 7 x 3 = 21.