Prove correctess of recurrence result by mathematical induction I have the following recurrence relation which I already solved using repeated substitution.

$T(n) = \begin{Bmatrix}
1 &amp; if &amp; n = 1\\ 
4T(\frac{n}2) + n &amp; if &amp; n &gt;= 2
\end{Bmatrix} where$ $n=2^k$

The result I get is $T(n) = 2n^2 - n$
Let´s suppose my result is correct. Now, I am asked to prove the correctness of the result using mathematical induction.

Base case is easy:
For $n = 1$ we have $T(1) = 2(1^2) - 1 = 2 - 1 = 1$
This result agrees with initial formulation of the problem above.

Now, comes the $Inductive$ $Hypothesis$ which I guess is as follows:
Assume this holds for $2 &lt;= k &lt;= n$
$4T(\frac{k+1}2) + (k+1) = 2(k+1)^2 - (k+1)$

Am I doing things correctly?
How would the $Induction$ $Step$ would be?
I will very much appreciate your feedback.
We prove by induction that for every $n\in\mathbb{N}$ it holds that $T(n)=2n^2-n$.
The inductive hypothesis is that for all $1\le k&lt;n$ we have:

$T(k)=2k^2-k$.

We proceed to show that $T(n)=2n^2-n$.

By definition, $T(n)=4T\left(\frac{n}{2}\right)+n$. Since $\frac{n}{2}&lt;n$, we can use our inductive hypothesis and obtain $T\left(\frac{n}{2}\right)=2\left(\frac{n}{2}\right)^2-\frac{n}{2}$. After substituting this back into $T(n)$ we have:

$T(n)=4\left(\frac{n^2}{2}-\frac{n}{2}\right)+n=2n^2-n$.