Having trouble find sum of this equation which has exponents. Can someone please help me solve this My question is:

Find the value of 
  $$\sum\limits_{j=1}^{100}(3^j+3\cdot 2^j)$$
   Leave answer as powers of $2$ and $3$.

I've really tried to think of a way to solve this but cant seem to find one. Could someone please show the steps to how they would solve this.

Since ${\sum_{j=0}^{n}ar^j=a\left(\frac{1-r^{n+1}}{1-r}\right)}\quad r\ne 1$, it follows
\begin{align}
\sum_{j=1}^{100}\left(3^j+3\cdot 2^j\right)&=\sum_{j=1}^{100}3^j+3\cdot\sum_{j=1}^{100}2^j\\
&=3\cdot\sum_{j=0}^{99}3^j+3\cdot2\cdot\sum_{j=0}^{99}2^j\\
&=3\left(\frac{1-3^{100}}{1-3}\right)+3\cdot 2\cdot\left(\frac{1-2^{100}}{1-2}\right)\\
&=\frac{3}{2}\left(3^{100}-1\right)+3\cdot 2\left(2^{100}-1\right)
\end{align}
