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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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prove that J[i] is an euclidean ring
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Assume E is the event that a randomly generated bit string of length 4 starts with a 1 and F is the event that this bit string consists of an even number of 1's. Are E and F indepe
Using the definition of the definite integral calculate the following. ∫ 0 2 x 2 + 1dx Solution Firstly,
2/6
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Formulas
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