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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
It is the last case that we need to take a look at. Throughout this section we will look at solutions to the system, x?' = A x? Here the eigenvalues of the matrix A are compl
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I need help with this question: Find the probability that two quarters and a nickel are chosen without replacement from a bag of 8 quarters and 12 nickles.
example and about this
1/cos(x-a)cos(x-b)
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