I have given $$ T(n)= \begin{cases} T(n/3)+T(2n/3)+n,\quad &n>1 \\ 1, \quad &n=1 \end{cases} $$ I tried it again and again but couldn't think beyond, $$ T(n)=T(n/27)+3T(2n/27)+3T(4n/27)+T(8n/27) \\ +n/3+2n/3+n/9+8n/9+n $$ What would be next step, if I write $27$ as $3^3$ or $3^k$?
2026-03-25 18:43:41.1774464221
Solving recurrences using substitution method
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