Wyatt claims that _ is equivalent to _ Which statement about his claim is true in every aspect?

Answer:The statement that is true in every aspect regarding his claim is:     False, because when combined in this manner, the alternating signs of the series are lost ( Wyatt ignored the negative in the first factor)Step-by-step explanation: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]