Proving a recurrence relation is satisfied by a order The question is to give a proof that shows that the recurrence relation
$$T(n) = mT(n/2)+an^2 $$ is satisfied by $$T(n) = O(n^{log m})$$ for $$m>4$$
The only technique we have learned is iteration/substitution. 
So I continued substituting 3 times until I got $$T(n) = m^4T(n/2^4)+a(m(n/2^3)^2 + m(n/2^2)^2+m(n/2)^2 + n^2) $$
Generally what we do now is we take the series at the end and try to see if it fits a usual series and convert and then find when $$n/2^k = 1$$ which gives us $$k=log_2n$$ and $$n = 2^k$$
I think the variable m and the m > 4 is throwing me off.  
I know there are answers that use the master theorem etc. but seeing as I've never learned those I want to stick to doing it in the way that we were taught. 
 A: You can prove by induction that:
\begin{align}
T(n) &= m^kT\left(\frac{n}{2^k}\right)+an^2\sum_{i=0}^{k-1}\left(\frac{m}{4}\right)^i\\
&=m^kT\left(\frac{n}{2^k}\right)+an^2\left(\frac{1 - \left(\frac{m}{4}\right)^k}{1-\frac{m}{4}}\right)\\
&=m^kT\left(\frac{n}{2^k}\right) + an^2\left(\frac{4^k-m^k}{4-m}\right)\frac{\frac{1}{4^k}}{\frac{1}{4}} \\
&=m^kT\left(\frac{n}{2^k}\right) + an^2\left(\frac{4^k-m^k}{4-m}\right)\frac{4}{4^k}
\end{align}
So for $k = \log_2(n)$, we reach the base case $T(1)$, usually assumed to be a constant $c$. 
\begin{align}
T(n) &= m^{\log_2(n)}c+an^2\left(\frac{4^{\log_2(n)}-m^{\log_2(n)}}{4-m}\right)\frac{4}{4^{\log_2(n)}} \\
&=m^{\log_2(n)}c + an^2\frac{n^2 - m^{\log_2(n)}}{4-m}\frac{4}{n^2}\\
&= m^{\log_2(n)}c + 4a\frac{n^2 - m^{\log_2(n)}}{4-m} \\
&= \left(c + \frac{4a}{m-4}\right)m^{\log_2(n)} - \frac{4a}{m-4}n^2
\end{align}
Since $m^{\log_2(n)} = 2^{\log_2(m)\log_2(n)}=n^{\log_2(m)}$, and $\log_2(m) > 2$ (because $m> 4$), we have:
$$T(n)= O\left(n^{\log(m)}\right)$$
as desired.
