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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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* 2^(1/2)*4^(1/8)*8^(1/16)*16^(1/32) =
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ABC is a right-angled isosceles triangle, right-angled at B. AP, the bisector of ∠BAC, intersects BC at P. Prove that AC 2 = AP 2 + 2(1+√2)BP 2 Ans: AC = √2AB (Sinc
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Ashow that sec^2x+cosec^2x cannot be less than 4
3x2 +7x +4
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