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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
what is the differeance in between determinate and matrix .
Ratio of successes in 5 independent trials to the probability of successes in two independent trials is 1/4. What is the probability of 4 successes in 6 independent trials?
matrix of [1 4 ] [a b]=4/9
Illustration : Solve the following IVP. Solution: First get the eigenvalues for the system. = l 2 - 10 l+ 25 = (l- 5) 2 l 1,2 = 5 Therefore, we got a
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