T(n) = T(n/3) + cn
T(3) = 3
Solving the recurrence relation by substitution.
But before that let n = 3$^{k}$ for easy calculation.
T(n) = T(n/3) + cn
= T(n/9) + cn/3 + cn
= T(n/27) + cn/9 + cn/3 + cn
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.
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= T($\frac{n}{3^{k-1}}$) + $\frac{cn}{3^{k-2}}$ + $\frac{cn}{3^{k-3}}$ + .... + $\frac{cn}{3^{k-k}}$
= T(3) + [$\frac{cn}{3^{k-2}}$ + $\frac{cn}{3^{k-3}}$ + .... + $\frac{cn}{3^{k-k}}$]
= 3 + cn [$\frac{1-\frac{1}{3^{k-1}}}{1-\frac{1}{3}}$]
= 3 + cn [$\frac{3^{k-1}-1}{\frac{2}{3}\times 3^{k-1}}$]
= 3 + (3c/2) n [$\frac{\frac{n}{3}-1}{\frac{n}{3}}$]
= 3 + (9c/2) [${\frac{n}{3}-1}$]
= $\Theta (n)$