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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
Q. Find the number of ways three letter "words" can be chosen from the alphabet if none of the letters can be repeated? Solution: There are 26 ways of choosing the first lett
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Identify the surface for each of the subsequent equations. (a) r = 5 (b) r 2 + z 2 = 100 (c) z = r Solution (a) In two dimensions we are familiar with that this
Equations in linear algebra and matrices What is Equations in linear algebra and matrices?
Objectives After going through this unit, you should be able to 1. explain the processes involved ih addition and subtraction; 2. plan and execute activities that woul
X^2 – y^2 – 2y - 1
You don''t have to give me the answer. I just want to know HOW to do it. In a set of 400 ACT scores where the mean is 22 and the standard deviation is 4.5, how many scores are ex
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