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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
ABCD is a rectangle. Δ ADE and Δ ABF are two triangles such that ∠E=∠F as shown in the figure. Prove that AD x AF=AE x AB. Ans: Consider Δ ADE and Δ ABF ∠D = ∠B
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If arg (a/b) = pi/2, then find the value of ((a+b)/(a-b)) where a,b are complex numbers. Ans) Arg (a/b) =Pi/2 Tan-1 (a/b)= Pi/2 A/B = tanP/2 ,therefore a/b=infinity.
how do you solve simultaneous equation?
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