Sea Empire: Winter lords

"Embark on a conquest of nearby islands in Sea Empire: Winter Lords by building a formidable fleet. This game boasts simple graphics and offers various ship options."<|endoftext|><|endoftext|> # 2006 AMC 12B Problems/Problem 1. (Redirected from 2006 AMC 12B Problems / Problem 1) ## Problem. What is the value of $2^{-3}+3^{-3}+4^{-3}+...+2006^{-3}+2007^{-3}$? $\text {(A) } \frac{2007}{2^3} \qquad \text {(B) } \frac{2007^2}{2^3} \qquad \text {(C) } \frac{2007^2+2007}{2^3} \qquad \text {(D) } \frac{2007^2+2007}{2^6} \qquad \text {(E) } \frac{2007^2+2007}{2^7}$ ## Solution. We can rewrite the expression as \begin{align*} 2^{-3}+3^{-3}+4^{-3}+...+2006^{-3}+2007^{-3} &= \left(2^{-3}+3^{-3}+4^{-3}+...+2006^{-3}\right) + 2007^{-3} \\ &= \left(\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{2006^3}\right) + \frac{1}{2007^3} \\ &= \frac{1}{8}+\frac{1}{27}+\frac{1}{64}+...+\frac{1}{2006^3} + \frac{1}{2007^3} \\ &= \left(\frac{1}{2^3}\right)^3+\left(\frac{1}{3^3}\right)^3+\left(\frac{1}{4^3}\right)^3+...+\left(\frac{1}{2006^3}\right)^3 + \left(\frac{1}{2007^3}\right)^3 \\ &= \left(\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{2006^3}+\frac{1}{2007^3}\right)^3 \\ &= \left(\frac{1}{8}+\frac{1}{27}+\frac{1}{64}+...+\frac{1}{2006^3}+\frac{1}{2007^3}\right)^3 \\ &= \left(\frac{1}{8}+\frac{1}{27}+\frac