First of all, we can factor the 2 out of the sum:
[tex]\displaystyle \sum_{n=0}^6 2\left(\dfrac{4}{3}\right)^n = 2\sum_{n=0}^6 \left(\dfrac{4}{3}\right)^n[/tex]
Now, you can use the result
[tex]\displaystyle \sum_{n=0}^k a^n = \dfrac{a^{k+1}-1}{a-1}[/tex]
So, in your case, we have
[tex]\displaystyle \sum_{n=0}^6 \left(\dfrac{4}{3}\right)^n = \dfrac{\left(\frac{4}{3}\right)^7-1}{\frac{4}{3}-1}=\dfrac{\frac{4^7-3^7}{3^7}}{\frac{1}{3}}=3\cdot \dfrac{4^7-3^7}{3^7} = \dfrac{4^7-3^7}{3^6}[/tex]
And finally, we have
[tex]\displaystyle 2\sum_{n=0}^6 \left(\dfrac{4}{3}\right)^n = 2\cdot\dfrac{4^7-3^7}{3^6} = \dfrac{28394}{729}\approx 38.95[/tex]
