The statement that is true in every aspect regarding his claim is:
The series is given as:
[tex]\sum_{n=0}^{3} (-\dfrac{1}{3})^n\cdot 9[/tex]
which could also be written by:
[tex]\sum_{n=0}^{3} (-\dfrac{1}{3})^n\cdot 9=\sum_{n=0}^{3} (-\dfrac{1}{3})^n\cdot 3^2\\\\\\\sum_{n=0}^{3} (-\dfrac{1}{3})^n\cdot 9=\sum_{n=0}^{3} (-1)^n(\dfrac{1}{3})^n\cdot 3^2[/tex]
i.e.
[tex]\sum_{n=0}^{3} (-\dfrac{1}{3})^n\cdot 9=\sum_{n=0}^{3} (-1)^n\cdot 3^{-n}\cdot 3^2\\\\\\\sum_{n=0}^{3} (-\dfrac{1}{3})^n\cdot 9=\sum_{n=0}^{3} (-1)^n\cdot 3^{2-n}[/tex]
which is not equivalent to: [tex]\sum_{n=0}^{3} 3^{2-n}[/tex]
Since, on expanding the actual series we get the sum as:
[tex]\sum_{n=0}^{3} (-\dfrac{1}{3})^n\cdot 9=(-\dfrac{1}{3})^0\cdot 9+(-\dfrac{1}{3})^1\cdot 9+(-\dfrac{1}{3})^2\cdot 9+(-\dfrac{1}{3})^3\cdot 9\\\\\\=9-\dfrac{1}{3}\times 9+\dfrac{1}{3^2}\times 9-\dfrac{1}{3^3}\times 9\\\\\\=9-3+1-\dfrac{1}{3}\\\\\\=\dfrac{22}{3}[/tex]
Now, the expansion of:
[tex]\sum_{n=0}^{3} 3^{2-n}[/tex] is:
[tex]\sum_{n=0}^{3} 3^{2-n}=3^2+3^1+3^0+3^{-1}\\\\\\=9+3+1+\dfrac{1}{3}=\dfrac{40}{3}[/tex]
