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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
"Standard" trig equation: Now we need to move into a distinct type of trig equation. All of the trig equations solved to this point were, in some way, more or less the "standard"
4.2^2x+1 - 9.2^x + 1=0
Example 1 Add 4x 4 + 3x 3 - x 2 + x + 6 and -7x 4 - 3x 3 + 8x 2 + 8x - 4 We write them one below the other as shown below.
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what is 1/3 + 2/9 equal
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