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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
In this case we will require deriving a new formula for variation of parameters for systems. The derivation now will be much simpler than the when we first noticed variation of pa
4.2^2x+1 - 9.2^x + 1=0
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Cartesian product - situations in which the total number of ordered pairs (or triples, or ...) are do be found. (e.g., if Hari makes 'dosas' of 3 different sizes, with 4 different
A circular disc of 6 cm radius is divided into three sectors with central angles 1200, 1500,900. What part of the circle is the sector with central angles 1200. Also give the ratio
/100*4500/12
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Hi I have a maths question related to construction as its a construction management course...i could send some example sheets too...could it be done?
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