Consider the following sequence: $a_j = 4a_{j-1} - 4a_{j-2}$
$a_0=0;\ a_1 = 1$

For all $j\geqslant2$, come up with a general formula for the term $a_j$. 
  Use mathematical induction to prove your claim.

I have calculated the first few terms of this sequence, and determined from them that $a_j = j · 2^{j-1}$.
However, I am very bad at induction (some might even say weak at it) and would appreciate if someone could help me prove that this is in fact the case.

In your case, since your linear recurrence relation uses $2$ smaller subscripts in the definition for $a_j$, you want to use strong induction to prove that
$$a_j = j\left(2^{j-1}\right), \; j \ge 0 \tag{1}\label{eq1A}$$
In this situation, you need to verify $2$ base cases. Although the question asks you prove \eqref{eq1A} holds for $j \ge 2$, it actually holds starting at $0$. As such, you can start there to first verify it works for $j = 0$ and $j = 1$. This is true since $0 = 0(2^{-1})$ and $1 = 1(2^{0})$, respectively.
Next, assume for some $k \ge 1$ that \eqref{eq1A} holds for all $j \le k$. For $j = k + 1$, you have
$$\begin{equation}\begin{aligned}
a_{k + 1} & = 4a_{k} - 4a_{k - 1} \\
& = 4k\left(2^{k-1}\right) - 4(k-1)\left(2^{k-2}\right) \\
& = 2k\left(2^{k}\right) - (k-1)\left(2^{k}\right) \\
& = 2^{k}(2k - (k - 1)) \\
& = (k+1)2^{k}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
This shows \eqref{eq1A} also holds for $j = k + 1$. Thus, by strong induction, \eqref{eq1A} holds for all $j \ge 0$.
