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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
Solve 2 ln (√x) - ln (1 - x ) = 2 . Solution: Firstly get the two logarithms combined in a single logarithm. 2 ln (√x) - ln (x - l) = 2 ln ((√x) 2 ) ln (1 - x ) = 2
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if 1/x+2, 1/x+3, 1/x+5 are in AP find x Ans 1/x+2,1/x+3, 1/x+5 are in AP find x. 1/x+3 - 1/x+2 = 1/x+5-1/x+3 => 1/x 2 +5x+6 = 2/ x 2 +8x +15 => On solving we get x
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