How to solve recursion T(n)=T(n/2)+T(n/3)+n?
This is a straightforward application of the Akra-Bazzi theorem.

You can also show that $T(n) = \Theta(n)$ by induction. First, clearly $T(n) \geq n$. In the other direction, if we have shown that $T(m) \leq 6m$ for $m &lt; n$, then
$$
T(n) \leq 6(n/2 + n/3) + n = 6n.
$$
Therefore as long as the base case holds, and if we ignore the fact that $n/2,n/3$ are not necessarily integers, we can prove inductively that $T(n) \leq 6n$.