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Solve the Recurrence Relation T(n) = 2T(n/3) + 2, T(1) = 1

Solve the Recurrence Relation T(n) = 2T(n/3) + 2, T(1) = 1

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Answer #1

Given, T(n) = 2*T(n/3) + 2 and T(1) = 1

Now, T(3) = 2*T(3/3) + 2 = 2*T(1) + 2 = 2*1 + 2 = 22

T(32) = T(9) = 2*T(9/3) + 2 = 2*T(3) + 2 = 2*[2*2] + 2 = 23 + 2

T(33) = T(27) = 2*T(27/3) + 2 = 2*(23 + 2) + 2 = 24 + 22 + 2

T(34) = T(81) = 2*T(81/3) + 2 = 2*(24 + 22 + 2) + 2 = 25 + 23 + 22 + 2

T(35) = T(243) = 2*T(243/3) + 2 = 2*(25 + 23 + 22 + 2) + 2 = 26 + 24 + 23 + 22 + 2

Proceeding in this way we get, T(3n) = 2n+1 + 2n-1 + 2n-2 + 2n-3 + ......... + 23 + 22 + 2.

Therefore, the required solution is, T(3n) = 2n+1 + 2n-1 + 2n-2 + 2n-3 + ......... + 23 + 22 + 2.

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