Question

Answer 1

a)

\(A=2^0+2^1+2^2+2^3+...+2^{2010}\\ 2A=2\cdot\left(1+2+2^2+...+2^{2010}\right)\\ 2A=2+2^2+2^3...+2^{2011}\\ 2A-A=\left(2+2^2+2^3+...+2^{2011}\right)-\left(1+2+2^2+...+2^{2010}\right)\\ A=2^{2011}-1\) 

b) \(B=1-3+3^2+...+3^{100}\)

\(B=3^2+3^3+...+3^{100}+\left(-2\right)\)

\(3B=3^3+3^4+...+3^{101}+\left(-2\right)\cdot3\)

\(3B-B=\left[3^3+3^4+...+3^{101}+\left(-6\right)\right]-\left[3^2+...+3^{100}+\left(-2\right)\right]\)

\(2B=\left(3^3-3^3\right)+\left(3^4-3^4\right)+...+\left(3^{101}-6-3^2+2\right)\)

\(2B=3^{101}-13\)

\(B=\dfrac{3^{101}-13}{2}\)