Question

 \(A=3^0-3^1+3^2-...-3^9+3^{10}\)

Answer 1

\(A=3^0-3^1+3^2-\cdot\cdot\cdot-3^9+3^{10}\)

\(3A=3^1-3^2+3^3-\cdot\cdot\cdot-3^{10}+3^{11}\)

\(3A+A=3^1-3^2+3^3-\cdot\cdot\cdot-3^{10}+3^{11}+3^0-3^1+3^2-\cdot\cdot\cdot-3^9+3^{10}\)

\(4A=3^{11}+1\)

\(\Rightarrow A=\dfrac{3^{11}+1}{4}\)

#Urushi☕

Answer 2

\(A=3^0-3^1+3^2-...-3^9+3^{10}\)
\(3A=3-3^2+3^3-...+3^{11}\)

\(3A+A=\left(3-3^2+3^3-...+3^{11}\right)+\left(1-3+3^2-...+3^{10}\right)\)

\(4A=3-3^2+3^3-...+3^{11}+1-3+3^2-...+3^{10}\)

\(4A=\left(3-3\right)+\left(3^2-3^2\right)+...+\left(3^{11}+1\right)\)

\(4A=3^{11}+1\)

\(A=\dfrac{3^{11}+1}{4}\)

Answer 3

A = \(\dfrac{3^{11}+1}{4}\)