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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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3x+y=9 5x-y=7
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nc6:n-3c3=91:4
All numbers refer to exercises (and not "computer exercises") in Gallian. §22: 8, 16, 22, 24, 28, 36. In addition: Problem 1: Let a be a complex root of the polynomial x 6 +
equivalent decimal for 25%
37x7= multiply answer it.
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