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Solve the recurrence relation
T (K) = 2T (K-1), T (0) = 1
Ans: The following equation can be written in the subsequent form:
tn - 2tn-1 = 0
Here now successively replacing n by (n - 1) and then by (n - 2) and so on we obtain a set of equations.
The method is continued till terminating condition. Add these equations in such type of a way that all intermediate terms get cancelled. The equation can be rearranged as
Multiplying all the equations correspondingly by 20, 21, ..., 2n - 1 and then adding them together, we get
tn - 2nt0 = 0
or, tn = 2n
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If a differential equation does have a solution how many solutions are there? As we will see ultimately, this is possible for a differential equation to contain more than one s
Here we know x can only be 1 or -1. so if it is 1 ans is 2. if x is -1, for n even ans will be 2 if x is -1 and n is odd ans will ne -2. so we can see evenfor negative x also an
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